The formula !a&EGa cannot hold in any state since it is a contradiction. You can see that by unrolling EGa into a & EX EG a which then contradicts directly !a. Hence, S2 is simply false and cannot hold in any state.
EGa holds in state s0, but not in state s1. Consider now E(F!a & Era) which is equivalent to (EF!a)&(EGa). This holds in state s0 since the path s0->s1^omega satisfies F!a and the path s0^omega satisfies EGa. Hence, s0 satisfies (EF!a)&(EGa), but s1 does not. Finally, abbreviate this formula by a new variable b which is therefore a label of s0, but not of s1, and consider the formula EGb. It clearly holds in s0, but not in s1, and therefore S1 holds in s0.
