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' ? 
> 
> 
> 



About Tarski's axioms a and A' and around (1)
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.


