Solution of a problem of size I know that the set of all groups does not exist. It's the same in the case of the topological spaces, and the sets themselves. My question is: How the Tarki's axiom solve this problem?
I can't see how the universes of Grothendieck gives sense to the staments "set of all groups", and the same.
 A: It doesn't, it gets around it.
There is (in $\mathsf{ZF}$ at least) no "set of all groups," and since that's a theorem of $\mathsf{ZF}$ adding additional axioms isn't going to change that. However, we can look at the particular ways in which we're tempted to use a hypothetical "set of all groups" in practice, and ask whether there might be some actually-consistent-with-$\mathsf{ZF}$ (+etc.) proxy for that.
Now $\mathsf{ZF}$ has a very nice framework for developing such ideas - namely, the cumulative hierarchy. For each (limit, for niceness) ordinal $\alpha$ we can talk about the set of all groups of set-theoretic rank (this is unrelated to the group-theoretic notion of rank) $<\alpha$; this will be a genuine set, and in fact an element of $V_{\alpha+1}$. The idea now is that if $V_\alpha$ is "nice enough," this approximation to the naive thing we want will actually do all the specific things we want it to.
This is where Tarski comes in. Things are very nice indeed if $V_\alpha$ is a Tarski universe (= if $\alpha$ is a strongly inaccessible cardinal), and we wind up with the following basic theme:

Suppose there is a proper class of strongly inaccessible ordinals $\alpha$. Then for any particular application we care about, we can find a set $\mathfrak{S}$ which is "sufficiently like" the class of all groups for our purposes - specifically, we'll take $\mathfrak{S}$ to be the set of groups in $V_\alpha$ for $\alpha$ a "sufficiently large" strongly inaccessible cardinal.

