Given the following two statements:
$$(S_1)$$ : $$(q \vee p) \to (p \leftrightarrow \sim q)$$ is a tautology
$$(S_2)$$ : $$\sim q \wedge (\sim p \leftrightarrow q)$$ is a fallacy. Then:

Let us first write clearly what every symbol means. The symbol $$\vee$$ stands for “or”, $$\wedge$$ for “and”, $$\sim$$ for “not”, $$\to$$ for “if … then …”, and $$\leftrightarrow$$ for “if and only if” (biconditional). A statement is a tautology when it is true for every possible truth-value assignment of its component variables, while it is a fallacy when it is false for every assignment.

We have two compound statements:

$$S_1 :\;(q \vee p) \to (p \leftrightarrow \sim q)$$

$$S_2 :\;\sim q \wedge (\sim p \leftrightarrow q)$$

To test whether $$S_1$$ is a tautology, we enumerate all possible truth-values of $$p$$ and $$q$$. There are four possibilities: $$(p,q) = (T,T),\,(T,F),\,(F,T),\,(F,F).$$ For every pair we shall compute each sub-expression step by step.

Case 1: $$p=T,\;q=T$$

First, $$q \vee p = T \vee T = T.$$

Next, $$\sim q = \sim T = F,$$ so $$p \leftrightarrow \sim q = T \leftrightarrow F = F$$ because a biconditional is true only when both sides have the same truth value.

Finally, $$(q \vee p) \to (p \leftrightarrow \sim q) = T \to F = F,$$ since “true implies false” is false.

Case 2: $$p=T,\;q=F$$

Here, $$q \vee p = F \vee T = T.$$

Also, $$\sim q = \sim F = T,$$ hence $$p \leftrightarrow \sim q = T \leftrightarrow T = T.$$

Thus $$(q \vee p) \to (p \leftrightarrow \sim q) = T \to T = T.$$

Case 3: $$p=F,\;q=T$$

We obtain $$q \vee p = T \vee F = T.$$

Further, $$\sim q = \sim T = F,$$ so $$p \leftrightarrow \sim q = F \leftrightarrow F = T.$$

Consequently, $$(q \vee p) \to (p \leftrightarrow \sim q) = T \to T = T.$$

Case 4: $$p=F,\;q=F$$

Now $$q \vee p = F \vee F = F.$$

Whenever the antecedent of an implication is false, the whole implication is automatically true, because $$F \to X = T$$ for any $$X$$. Therefore we do not even need to calculate the biconditional; still, for completeness we note $$\sim q = T$$ and $$p \leftrightarrow \sim q = F \leftrightarrow T = F.$$ The implication is $$F \to F = T.$$

Collecting the truth values of $$S_1$$ we have: $$F,\;T,\;T,\;T.$$ Because one row (the very first) gives false, $$S_1$$ is not a tautology. So statement $$S_1$$ is incorrect.

Next we examine $$S_2 = \sim q \wedge (\sim p \leftrightarrow q)$$ to see whether it is a fallacy. Again we test all four assignments.

Case 1: $$p=T,\;q=T$$

Compute $$\sim q = F.$$

Compute $$\sim p = F,$$ so $$\sim p \leftrightarrow q = F \leftrightarrow T = F.$$

Hence $$\sim q \wedge (\sim p \leftrightarrow q) = F \wedge F = F.$$

Case 2: $$p=T,\;q=F$$

We get $$\sim q = T.$$

Also $$\sim p = F,$$ and $$\sim p \leftrightarrow q = F \leftrightarrow F = T.$$

Therefore $$\sim q \wedge (\sim p \leftrightarrow q) = T \wedge T = T.$$ This row gives true.

Case 3: $$p=F,\;q=T$$

Here $$\sim q = F.$$

Then $$\sim p = T,$$ and $$\sim p \leftrightarrow q = T \leftrightarrow T = T.$$

Thus $$\sim q \wedge (\sim p \leftrightarrow q) = F \wedge T = F.$$

Case 4: $$p=F,\;q=F$$

Now $$\sim q = T.$$

Also $$\sim p = T,$$ giving $$\sim p \leftrightarrow q = T \leftrightarrow F = F.$$

Hence $$\sim q \wedge (\sim p \leftrightarrow q) = T \wedge F = F.$$

The truth values obtained for $$S_2$$ are $$F,\;T,\;F,\;F.$$ Because at least one row (the second) is true, the statement is not always false; therefore it is not a fallacy. Hence statement $$S_2$$ is also incorrect.

We have shown that $$S_1$$ is not a tautology and $$S_2$$ is not a fallacy. Thus both statements asserted in the question are wrong.

Hence, the correct answer is Option A.