Both don't exist. So first I'll prove that set S1 doesn't exist and then I'll extend this proof to show that S2 doesn't exist.
Claim 1: S1 does not exist.
Proof: Let's assume that it does exist. First let's understand what it means - a set containing all those sets which don't contain themselves. Now let's ask this question - does S1 contain itself? There are 2 possibilities here 'YES' or 'NO'. If the answer is Yes, you mean to say that S1 does contain itself - it's a contradiction because the set can only contain those sets which don't contain themselves. But then if the answer is a NO, then it should be yes otherwise it will be a contradiction. So that is the reason why S1 cannot exist - and this is called Russell's paradox.
Claim 2: S2 does not exist.
Proof:
Method 1. Again let's start on a positive note assuming that S2 exists. So again let's visualize: this set is the universal set- the set which contains "all" the sets. If you say S2 exists, then it should contain all the sets you can possibly guess. Now I'll ask you this question - does the so proclaimed universal set contain the set S1? Just now we showed that this set cannot exist - and thus doesn't belong to S2. So S2 cannot be called the 'universal' set as I've shown a set which doesn't belong to it. Hence this proves the non existence of S2 aswell.
Hence the answer is option (C).
Method 2. We know that in general, the cardinality of a set is smaller than its power set. So let's assume that the universal set exists. Since the universal set contains literally everything, it should also contain its own power set, which makes $2^S$ a subset of the 'assumed' universal set $S$. And we also know that if a set $A$ is a subset of a set $B$, then $|A| <= |B|$, it in turn implies $|2^S|$ is smaller than $|S|$, which violates the above guideline - power set of a set can't be smaller in cardinality than the set and thus a contradiction. Hence the universal set cannot be real.
