A Grothendieck universe is known in set theory as the set Vκ for a (strongly) inaccessible cardinal κ. They are exactly the same thing. Thus, the existence of a Grothendieck universe is exactly equivalent to the existence of one inaccessible cardinal. These cardinals and the corresponding universes have been studied in set theory for over a century.
The Grothendieck Universe axiom (AU) is the assertion that every set is an element of a universe in this sense. Thus, it is equivalent to the assertion that the inaccessible cardinals are unbounded in the cardinals. In other words, that there is a proper class of inaccessible cardinals. This is the axiom you sought, which is exactly equivalent to AU. In this sense, the axiom AU is a statement in set theory, having nothing necessarily to do with category theory.
The large cardinal axioms are fruitfully measured in strength not just by direct implication, but also by their consistency strength. One large cardinal property LC1 is stronger than another LC2 in consistency strength if the consistency of ZFC with an LC1 large cardinal implies the consistency of ZFC with an LC2 large cardinal.
Measured in this way, the AU axiom has a stronger consistency strength than the existence of any finite or accessible number of inaccessible cardinals, and so one might think it rather strong. But actually, it is much weaker than the existence of a single Mahlo cardinal, the traditional next-step-up in the large cardinal hierarchy. The reason is that if κ is Mahlo, then κ is a limit of inaccessible cardinals, and so Vκ will satisfy ZFC plus the AU axiom. The difference between AU and Mahloness has to do with the thickness of the class of inaccessible cardinals. For example, strictly stronger than AU and weaker than a Mahlo cardinal is the assertion that the inaccessible cardinals form a stationary proper class, an assertion known as the Levy Scheme (which is provably equivconsistent with some other interesting axioms of set theory, such as the boldface Maximality Principle, which I have studied a lot). Even Mahlo cardinals are regarded as rather low in the large cardinal hierarchy, far below the weakly compact cardinals, Ramsey cardinals, measurable cardinals, strong cardinals and supercompact cardinals. In particular, if δ is any of these large cardinals, then δ is a limit of Mahlo cardinals, and certainly a limit of strongly inaccessible cardinals. So in particular, Vδ will be a model of the AU axiom.
Rather few of the large cardinal axioms imnply AU directly, since most of them remain true if one were to cut off the universe at a given inaccessible cardinal, a process that kills AU. Nevertheless, implicit beteween levels of the large caridnal hiearchy are the axioms of the same form as AU, which assert an unbounded class of the given cardinal. For example, one might want to have unboundedly many Mahlo cardinals, or unboundedly many measurable cardinals, and so on. And the consistency strength of these axioms is still below the consistency strength of a single supercompact cardinal. The hierarchy is extremely fine and intensely studied. For example, the assertion that there are unboundedly many strong cardinals is equiconsistent with the impossiblity to affect projective truth by forcing. The existence of a proper class of Woodin cardinals is particularly robust set-theoretically, and all of these axioms are far stronger than AU.
There are natural weakenings of AU that still allow for almost all if not all of what category theorists do with these universes. Namely, with the universes, it would seem to suffice for almost all category-theoretic purposes, if a given universe U were merely a model of ZFC, rather than Vκ for an inaccessible cardinal κ. The difference is that U is merely a model of the Power set axiom, rather than actually being closed under the true power sets (and similarly using Replacement in place of regularity). The weakening of AU I have in mind is the axiom that asserts that every set is an element of a transitive model of ZFC. This assertion is strictly weaker in consistency strength thatn even a single inaccessible cardinal. One can get much lower, if one weakens the concept of universe to just a fragment of ZFC. Then one could arrive at a version of AU that was actually provable in ZFC, but which could be used for most all of the applications in cateogory theory to my knowledge. In this sense, ZFC itself is a kind of large cardinal axiom relative to the weaker fragments of ZFC.
Yes. Assume ZFC. If there is a proper class of inaccessible cardinals, then Tarski's Axiom A holds because whenever $\kappa$ is inaccessible, the rank initial segment $V_\kappa$ of $V$ is a Tarski set. Conversely, if Tarski's Axiom A holds then for every set $x$ there is a Tarski set $y$ with $x \in y$. We will show that $|y|$ is an inaccessible cardinal greater than $|x|$, proving the existence of a proper class of inaccessible cardinals.
To show that the cardinality $\kappa$ of $y$ is a strong limit cardinal, given $\zeta < \kappa$ we take a subset $z$ of $y$ of size $\zeta$. We have $z \in y$ because $y$ contains its small subsets. Then we have $\mathcal{P}(z) \in y$ because $y$ is closed under the power set operation. Finally $\mathcal{P}(\mathcal{P}(z)) \subset y$ because $y$ contains all subsets of its elements. This shows that $2^{2^{\zeta}} \le \kappa$ and therefore that $2^{\zeta} < \kappa$.
To show that the cardinality $\kappa$ of $y$ is regular, notice that if $\kappa$ is singular then by the closure of $y$ under small subsets we can get a family of $\kappa^{cof(\kappa)}$ many distinct sets in $y$, contradicting the fact that $\kappa^{cof( \kappa)} > \kappa$ (which is an instance of Koenig's Theorem.)
Best Answer
The answer to question 1 is affirmative for GU and IN but negative for TA. That is, the proper class formulation of TA is not equivalent to TA, unless both are inconsistent.
Asserting that every ordinal is below an inaccessible cardinal is clearly equivalent in ZF to asserting that the inaccessible cardinals form a proper class. And since the Grothendieck universes are exactly the sets $V_\kappa$ for an inaccessible cardinal $\kappa$, the theory GU over ZFC is equivalent to the assertion that there are a proper class of universes. So those cases are relatively straightforward, as you had guessed.
But the case of the Tarski axiom is different, since his universes are not transitive sets. In fact, if there is a single inaccessible cardinal $\kappa$, one can already form a proper class of Tarski sets. To see this, suppose that $\kappa$ is inaccessible and $x$ is any set. Build a Tarski set $U$ as follows:
That is, we perform the usual cumulative hiearchy, but starting with object {x} instead of nothing. The resulting set $U=U_\kappa$ contains {x} as an element, has size $\kappa$ and is closed under subsets, power sets and small unions, so it is a Tarski universe. (Note that if $y\in U_{\alpha+1}$, then all subsets of $y$ are also in $U_{\alpha+1}$, and so $P(y)$ is added on the next step. And {x} has only two subsets, which are both in $U$.) But for sufficiently large $x$, the resulting $U$ will not be transitive, since $x$ itself will never be added as an element. Furthermore, since $x$ was arbitrary, from a single inaccessible cardinal we can build a proper class of Tarski universes. And so the proper class formulation of TA will not be equivalent to TA, unless both theories are inconsistent, because if the existence of an inaccessible cardinal is consistent with ZFC, then it is consistent that there is exactly one such cardinal, by chopping the universe off at the second one. The resulting model will satisfy the proper class formulation of TA but not TA itself.