1. If ∀x(A(x)⇔B(x)) is true, then for every choice of x, A(x) implies B(x) and conversely.
Pick any x, then A(x) implies B(x), so, there exists an x such that A(x) implies B(x).
Assuming that hypothesis is true, we arrive at the conclusion being true, so S1 is a true statement.
2. Hypothesis is saying that there exists an x such that either A(x) or B(x) is true. If that is true, then for that particular x, either A(x) is true or B(x) is true.
That implies that there exists an x such that A(x) is true, or there exists an x such that B(x) is true. From hypothesis we arrive at the conclusion.
Converse part of S2 is obvious too. So S2 is true.