About Tarski's axioms a and A' and around (1) 1-In his article written in German "Über unerreichbare Kardinalzahlen" (On inaccessible cardinals), inside Fund. Math. 1938 (pages 68-89), Alfred Tarski states his axioms A and A' as follows.
Axiom A: "For every set x, there exists a set y satisfying the four following conditions:


*

*A1: x is a member element of y;

*A2: If z is a member element of y and t is included inside z, then t is a member element of y;

*A3: If z is a member element of y and t is  the set having as member elements exactly all sets u included inside z, then t is a member element of y;

*A4: If z is included inside y and if the sets z and y are equipotent, then z is a member element of y."


Axiom A':"For every set x, there exists a set y satisfying the four following conditions:


*

*A'1: x is included inside y;

*A'2: If z is a member element of y and t is a member element of z, then t is a member element of y;

*A'3: If z is a member element of y, there exists a set w that is a member element of y such that every set t that is included inside z is a member element of w;

*A'4: If z is included inside y and if no set included inside z is equipotent with y, then z is a member element of y."


Tarski asserts, without giving a proof, that axioms A and A' are equivalent.

QUESTION 1: Does anyone know about an explicit proof of the equivalance of A and A' ? 

 A: First, I note that you appear to be missing a not in A4,
and you should say that "if $z$ and $y$ are not
equipollant", for otherwise we could take $z=y$ and thereby
deduce $y\in y$, contrary to the Foundation axiom.
With this correction, both your axioms are equivalent in
ZFC to the assertion that there is a proper class of
inaccessible cardinals.
For the one direction, if there are such cardinals, then
for any set $x$ we may find an inaccessible cardinal
$\kappa$ such that $x\in V_\kappa$, and take
$y=V_\kappa=H_\kappa$ to fulfill either $A$ or $A'$, which
is easily verified.
Conversely, assume axiom A. Let $x=\alpha$ be an ordinal
and let $y$ arise as in axiom A. Let $\kappa=|y|$. As every
subset of $\alpha$ is in $y$, it follows that
$\alpha\lt\kappa$. If $\beta\lt\kappa$, then there is
subset $z\subset y$ of size $\beta$, and this is an element
of $y$ by A4. We also know $P(z)\subset y$ and $P(z)\in Y$,
so $P(P(z))\subset y$, so $2^\beta\lt\kappa$. Thus,
$\kappa$ is a strong limit. For regularity, suppose that
$\kappa$ singular with cofinality $\gamma\lt\kappa$. Thus,
$y$ is the union of $\gamma$ many subsets, each of size
less than $\kappa$. These subsets are elements of $y$, and
all their subsets are also in $y$. But every subset of $y$
is determined by a similar $\gamma$ sequence of elements of
$y$, and so $y$ will have $2^\kappa$ many elements, a
contradiction. So $\kappa$ is an inaccessible cardinal
above $\alpha$, as desired.
For the other converse direction, assume axiom $A'$.
Consider $x=\alpha+1$, and get $y$ as in $A'$, and again
let $\kappa=|y|$. I claim that $\kappa\subset y$, for
otherwise the least ordinal $\beta$ not in $y$ would be
less than $\kappa$ in size and have all its subsets size
less than $\kappa$, and hence in $y$ by $A4'$. Thus,
$\kappa\subset y$. Now, for any $\gamma\lt\kappa$, every
subset of $\gamma$ is in $y$ and there is an element of $y$
with at least $2^\gamma$ many subsets, all in $y$, so
$2^\gamma\lt\kappa$. So $\kappa$ is a strong limit.
Regularity follows as before, and so $\kappa$ is an
inaccessible cardinal above $\alpha$.
Thus, since the axioms are both equivalent to the assertion that there is a proper class of inaccessible cardinals, they are also equivalent to each other.
Do you have some historical reason to study Tarski's treatment of inaccessibility? 
If not, I think you might find the contemporary accounts of large cardinals to be more appealing. You might look at Kanamori's book, The Higher Infinite.
If you wanted the equivalence in ZF rather than ZFC, or in
ZF-Foundation, then I wouuld have to think more carefully about it, but I will mention that this issue seems to be remarked on in Solovay's letter mentioned in
your previous question. In particular, without AC there are
competing inequivalent notions of inaccessibility.
