$S_1$ is false. The counter example is a very popular regular language - $(a+b)^*$
$S_2$ is true and at first glance, it looks like its prove must be subtle. But no, it is not difficult to notice that LHS will never have an empty string $\epsilon$, however, the RHS will always have an empty string. This one small difference on the two side of equality will never let them be equal to each other.
Thus, the correct answer is (d) S1 → False, S2 → True
HTH