Among the statements
(S1): $$(p \Rightarrow q) \lor ((\sim p) \wedge q)$$ is a tautology
(S2): $$(q \Rightarrow p) \Rightarrow ((\sim p) \wedge q)$$ is a contradiction

We begin by determining which of the two statements is true.

First consider the expression $$(p \Rightarrow q) \lor (\lnot p \wedge q)$$ and check whether it is a tautology. Recall that $$p \Rightarrow q \equiv \lnot p \lor q$$. Hence

since $$\lnot p \wedge q$$ is absorbed by $$\lnot p \lor q$$. Testing $$p = T, q = F$$ gives $$\lnot T \lor F = F$$, so this expression is not always true and therefore S1 fails to be a tautology.

Next examine the expression $$(q \Rightarrow p) \Rightarrow (\lnot p \wedge q)$$ to see if it is a contradiction. Noting that $$q \Rightarrow p \equiv \lnot q \lor p$$, we find

and testing $$p = F, q = T$$ yields $$\lnot F \wedge T = T$$. Since the expression can be true, S2 is not a contradiction.

For completeness, the following truth table verifies the values of S1 and S2 under all assignments of $$p$$ and $$q$$:

Since S1 is not a tautology (it fails when $$p = T, q = F$$) and S2 is not a contradiction (it holds when $$p = F, q = T$$), neither statement is true.

The correct answer is Option A: Neither (S1) nor (S2) is True.