Does the fact that, assuming the consistency of $ZFC$, no proof that the consistency of "$ZFC$ implies the consistency of '$ZFC$ + There exists a weakly inaccessible cardinal'" can be formulated in $ZFC$ (this paraphrased from the Wikipedia entry, "Inaccessible cardinal") imply that, given a model $M$ of $ZFC$ in which no weakly inaccessible cardinal exists, there can be no generic extension $M[G]$ of $M$ in which a weakly inaccessible cardinal was forced to exist?

I conjecture that the answer to this question is 'yes' due to the answers given to Ewen Delanoy to his mathstackexchange question, "Relation between inaccessible cardinals and $CH$." The answers were given by arjafi and Asaf Karagila. I quote the relevant paragraph of Asaf's answer first (my comments will be in square brackets throughout):

Do note, however, that the assumption of $CH$ is equiconsistent with its negation, as Godel's and Cohen's work show; while this is not true for the existence of large cardinals. The assertion that there exists a [weak] inaccessible is strictly stronger than the assertion that there are none [this asymmetry (I believe) implies that there is no notion of forcing that will produce a generic extension $M[G]$ of a model $M$ of $ZFC$ containing no weakly inaccessible cardinal such that $M[G]$ contains a weak inaccessible since forcing is essentially a relative consistency proof, i.e., that given a model $M$ of $ZFC$ ($CON(ZFC)$), there is a forcing extension $M[G]$ in which $\varphi$ holds ($CON($ZFC + ${\varphi}$))].

The reason for this asymmetry is described by arjafi in his answer:

This is where we have to hedge a bit. It is actually unknown whether any large cardinal assumptions are relatively consistent with $ZFC$. Furthermore, we cannot prove that large cardinals are relatively consistent with $ZFC$ without transcending $ZFC$ itself; that is proving the relative consistency from a metatheory stronger than $ZFC$.

So the following question remains (assuming the answer to my first question is 'yes'):

"What is this metatheory?" (Another way of possibly rephrasing this question is: "What strong base theory $T$ in the language of set theory (could be second-order or greater) which does not assume outright the existence of an inaccessible cardinal can have (say) $PRA$ proves "$CON$($T$ + There exists a weakly inaccessible cardinal) $\Leftrightarrow$ $CON$( $T$ + There is no weakly inaccessible cardinal)".) This state of affairs (I think) would allow one to have notions of forcing which, from a model $M$ of "$T$ + 'There is no weakly inaccessible cardinal'" one could construct a generic extension $M[G]$ satisfying "$T$ + 'There exists a weakly inaccessible cardinal'".

A natural candidate for $T$ might be $KM$ (Kelly-Morse) set theory, but as correctly argued in Joel David Hamkins' blogpost "Kelly-Morse set theory implies $CON(ZFC)$ and much more":

...the consistency strength of $KM$ lies strictly that of $ZFC$, above much of the iterated consistency hierarchy, but below that of $ZFC$ plus an inaccessible cardinal.

so $T$ $\ne$ $KM$.

Does the fact that, assuming the consistency of $ZFC$, no proof that the consistency of "$ZFC$ implies the consistency of '$ZFC$ + There exists a weakly inaccessible cardinal'" can be formulated in $ZFC$ (this paraphrased from the Wikipedia entry, "Inaccessible cardinal") imply that, given a model $M$ of $ZFC$ in which no weakly inaccessible cardinal exists, there can be no generic extension $M[G]$ of $M$ in which a weakly inaccessible cardinal was forced to exist?

I conjecture that the answer to this question is 'yes' due to the answers given to Ewen Delanoy to his mathstackexchange question, Relation between inaccessible cardinals and $CH$. The answers were given by arjafi and Asaf Karagila. I quote the relevant paragraph of Asaf's answer first (my comments will be in square brackets throughout):

Do note, however, that the assumption of $CH$ is equiconsistent with its negation, as Godel's and Cohen's work show; while this is not true for the existence of large cardinals. The assertion that there exists a [weak] inaccessible is strictly stronger than the assertion that there are none [this asymmetry (I believe) implies that there is no notion of forcing that will produce a generic extension $M[G]$ of a model $M$ of $ZFC$ containing no weakly inaccessible cardinal such that $M[G]$ contains a weak inaccessible since forcing is essentially a relative consistency proof, i.e., that given a model $M$ of $ZFC$ ($CON(ZFC)$), there is a forcing extension $M[G]$ in which $\varphi$ holds ($CON($ZFC + ${\varphi}$))].

The reason for this asymmetry is described by arjafi in his answer:

This is where we have to hedge a bit. It is actually unknown whether any large cardinal assumptions are relatively consistent with $ZFC$. Furthermore, we cannot prove that large cardinals are relatively consistent with $ZFC$ without transcending $ZFC$ itself; that is proving the relative consistency from a metatheory stronger than $ZFC$.

So the following question remains (assuming the answer to my first question is 'yes'):

"What is this metatheory?" (Another way of possibly rephrasing this question is: "What strong base theory $T$ in the language of set theory (could be second-order or greater) which does not assume outright the existence of an inaccessible cardinal can have (say) $PRA$ proves "$CON$($T$ + There exists a weakly inaccessible cardinal) $\Leftrightarrow$ $CON$( $T$ + There is no weakly inaccessible cardinal)".) This state of affairs (I think) would allow one to have notions of forcing which, from a model $M$ of "$T$ + 'There is no weakly inaccessible cardinal'" one could construct a generic extension $M[G]$ satisfying "$T$ + 'There exists a weakly inaccessible cardinal'".

A natural candidate for $T$ might be $KM$ (Kelly-Morse) set theory, but as correctly argued in Joel David Hamkins' blogpost "Kelly-Morse set theory implies $CON(ZFC)$ and much more":

...the consistency strength of $KM$ lies strictly that of $ZFC$, above much of the iterated consistency hierarchy, but below that of $ZFC$ plus an inaccessible cardinal.

so $T$ $\ne$ $KM$.

Does the fact that, assuming the consistency of $ZFC$, no proof that the consistency of "$ZFC$ implies the consistency of '$ZFC$ + There exists a weakly inaccessible cardinal'" can be formulated in $ZFC$ (this paraphrased from the Wikipedia entry, "Inaccessible cardinal") imply that, given a model $M$ of $ZFC$ in which no weakly inaccessible cardinal exists, there can be no generic extension $M[G]$ of $M$ in which a weakly inaccessible cardinal was forced to exist?

I conjecture that the answer to this question is 'yes' due to the answers given to Ewen Delanoy to his mathstackexchange question, "Relation between inaccessible cardinals and $CH$." The answers were given by arjafi and Asaf Karagila. I quote the relevant paragraph of Asaf's answer first (my comments will be in square brackets throughout):

Do note, however, that the assumption of $CH$ is equiconsistent with its negation, as Godel's and Cohen's work show; while this is not true for the existence of large cardinals. The assertion that there exists a [weak] inaccessible is strictly stronger than the assertion that there are none [this asymmetry (I believe) implies that there is no notion of forcing that will produce a generic extension $M[G]$ of a model $M$ of $ZFC$ containing no weakly inaccessible cardinal such that $M[G]$ contains a weak inaccessible since forcing is essentially a relative consistency proof, i.e., that given a model $M$ of $ZFC$ ($CON(ZFC)$), there is a forcing extension $M[G]$ in which $\varphi$ holds ($CON($ZFC + ${\varphi}$))].

The reason for this asymmetry is described by arjafi in his answer:

This is where we have to hedge a bit. It is actually unknown whether any large cardinal assumptions are relatively consistent with $ZFC$. Furthermore, we cannot prove that large cardinals are relatively consistent with $ZFC$ without transcending $ZFC$ itself; that is proving the relative consistency from a metatheory stronger than $ZFC$.

So the following question remains (assuming the answer to my first question is 'yes'):

"What is this metatheory?" (Another way of possibly rephrasing this question is: "What strong base theory $T$ in the language of set theory (could be second-order or greater) which does not assume outright the existence of an inaccessible cardinal can have (say) $PRA$ proves "$CON$($T$ + There exists a weakly inaccessible cardinal) $\Leftrightarrow$ $CON$( $T$ + There is no weakly inaccessible cardinal)". This state of affairs (I think) would allow one to have notions of forcing which, from a model $M$ of "$T$ + 'There is no weakly inaccessible cardinal'" one could construct a generic extension $M[G]$ satisfying "$T$ + 'There exists a weakly inaccessible cardinal'".

A natural candidate for $T$ might be $KM$ (Kelly-Morse) set theory, but as correctly argued in Joel David Hamkins' blogpost "Kelly-Morse set theory implies $CON(ZFC)$ and much more":

...the consistency strength of $KM$ lies strictly that of $ZFC$, above much of the iterated consistency hierarchy, but below that of $ZFC$ plus an inaccessible cardinal.

so $T$ $\ne$ $KM$.

Does the fact that, assuming the consistency of $ZFC$, no proof that the consistency of "$ZFC$ implies the consistency of '$ZFC$ + There exists a weakly inaccessible cardinal'" can be formulated in $ZFC$ (this paraphrased from the Wikipedia entry, "Inaccessible cardinal") imply that, given a model $M$ of $ZFC$ in which no weakly inaccessible cardinal exists, there can be no generic extension $M[G]$ of $M$ in which a weakly inaccessible cardinal was forced to exist?

I conjecture that the answer to this question is 'yes' due to the answers given to Ewen Delanoy to his mathstackexchange question, "Relation between inaccessible cardinals and $CH$." The answers were given by arjafi and Asaf Karagila. I quote the relevant paragraph of Asaf's answer first (my comments will be in square brackets throughout):

Do note, however, that the assumption of $CH$ is equiconsistent with its negation, as Godel's and Cohen's work show; while this is not true for the existence of large cardinals. The assertion that there exists a [weak] inaccessible is strictly stronger than the assertion that there are none [this asymmetry (I believe) implies that there is no notion of forcing that will produce a generic extension $M[G]$ of a model $M$ of $ZFC$ containing no weakly inaccessible cardinal such that $M[G]$ contains a weak inaccessible since forcing is essentially a relative consistency proof, i.e., that given a model $M$ of $ZFC$ ($CON(ZFC)$), there is a forcing extension $M[G]$ in which $\varphi$ holds ($CON($ZFC + ${\varphi}$))].

The reason for this asymmetry is described by arjafi in his answer:

This is where we have to hedge a bit. It is actually unknown whether any large cardinal assumptions are relatively consistent with $ZFC$. Furthermore, we cannot prove that large cardinals are relatively consistent with $ZFC$ without transcending $ZFC$ itself; that is proving the relative consistency from a metatheory stronger than $ZFC$.

So the following question remains (assuming the answer to my first question is 'yes'):

"What is this metatheory?" (Another way of possibly rephrasing this question is: "What strong base theory $T$ in the language of set theory (could be second-order or greater) which does not assume outright the existence of an inaccessible cardinal can have (say) $PRA$ proves "$CON$($T$ + There exists a weakly inaccessible cardinal) $\Leftrightarrow$ $CON$( $T$ + There is no weakly inaccessible cardinal)".) This state of affairs (I think) would allow one to have notions of forcing which, from a model $M$ of "$T$ + 'There is no weakly inaccessible cardinal'" one could construct a generic extension $M[G]$ satisfying "$T$ + 'There exists a weakly inaccessible cardinal'".

A natural candidate for $T$ might be $KM$ (Kelly-Morse) set theory, but as correctly argued in Joel David Hamkins' blogpost "Kelly-Morse set theory implies $CON(ZFC)$ and much more":

...the consistency strength of $KM$ lies strictly that of $ZFC$, above much of the iterated consistency hierarchy, but below that of $ZFC$ plus an inaccessible cardinal.

so $T$ $\ne$ $KM$.