I'm not sure I have the intended sense of the question, as there seem to be several ways to interpret it,
so let me give several replies. Please clarify if I've misunderstood.

- Perhaps you intend to ask: is there a formula
$\varphi(x)$ in the language of set theory such that
$L_\alpha\models\varphi(\kappa)$ if and only if $\kappa$ is
an uncountable cardinal in $V$? And is there a formula
$\psi(x)$ such that $L_\alpha\models\psi(\kappa)$ if and
only if $\kappa$ is inaccessible in $V$?

In this case, the answer is no. One way to see that we
should expect this is that by forcing we may make any set
countable (and hence also non-inacessible), but forcing
does not change the satisfaction relation of formulas in
$L$ or in $L_\alpha$, and so we may change the truth of the
right-hand-side of these proposed equivalences by moving to
a forcing extension, without changing the left-hand-side.
Thus, they cannot always be equivalent. But another direct
argument is that whenever $L_\alpha\models\varphi(\kappa)$,
then by the downward Lowenheim-Skolem theorem, there is in
$L$ a countable elementary substructure $X\prec L_\alpha$
containing $\kappa$, and the Mostowski collapse of $X$ is
isomorphic by the condensation principle to some $L_\beta$,
which will satisfy $\varphi(\kappa_0)$ for the image of
$\kappa$ under the collapse. Thus,
$L_\beta\models\varphi(\kappa_0)$, but $\kappa_0$ is
countable in $L$, violating the desired property.

- Perhaps you intend to ask: is there a formula
$\varphi(x)$ such that $L\models\varphi(\kappa)$ if and
only if $\kappa$ is uncountable in $V$ (and another formula
for inaccessibility).

In this case, the answer is that it depends on $V$. On the
one hand, if $V=L$, then there are such a formula, because
the property, $\kappa$ is uncountable,'' is expressible
in the first-order language of set theory, as the assertion
that there is no surjective function from $\omega$ to
$\kappa$. Similarly the property of being inaccessible is
expressible. The point is that it is consistent with ZFC
that the concepts of uncountable and inaccessible are in
agreement between $L$ and $V$.

But meanwhile, it is also consistent that there are no such
formulas. For example, one quick way to see this is that if
$0^\sharp$ exists, then all the Silver indiscernible
ordinals have the same first-order properties in $L$, and
so from the perspective of $L$, the cardinal $\aleph_1^V$
satisfies the same formulas as many countable ordinals.

But one needn't make the $0^\sharp$ assumption, and one can
do it equiconsistently with ZFC. The reason is that it is
equiconsistent with ZFC that there is a cardinal $\delta$
with $L_\delta\prec L$, expressed as a scheme in the
language with a constant for $\delta$. In such a model, we
may move to the forcing extension $L[G]$ collapsing
$\delta$ to $\omega$. In $L[G]$, all the uncountable
ordinals are above $\delta$, but by our $L_\delta\prec L$
hypothesis, for any ordinal above $\delta$ with a certain
property in $L$, there will be ordinals below $\delta$ with
that same property. But since these will all be countable
in $L[G]$, it violates the desired feature.

François pointed out in the comments below that you may have intended the question:

- Is it consistent that every ordinal that is definable in $L$ is countable in $V$?

The answer here is yes, since in the model $L[G]$ above, where $L_\delta\prec L$ and $G$ collapses $\delta$ to $\omega$, we have that every definable object of $L$ is in $L_\delta$, and these are all countable in $L[G]$.