Consider the two statements:
$$(S_1) : (p \rightarrow q) \vee (\sim q \rightarrow p)$$ is a tautology.
$$(S_2) : (p \wedge \sim q) \wedge (\sim p \vee q)$$ is a fallacy.
Then:

We have to examine two logical sentences.

First consider $$S_1 : (p \rightarrow q)\;\vee\;(\sim q \rightarrow p).$$

Before simplifying, we recall the basic implication equivalence:
For any propositions $$a$$ and $$b,$$ the implication $$a \rightarrow b$$ is equivalent to $$\sim a \;\vee\; b.$$

Applying this rule to each implication inside $$S_1,$$ we get

$$ (p \rightarrow q)=\; \sim p \;\vee\; q,\\ (\sim q \rightarrow p)=\; \sim (\sim q) \;\vee\; p=\; q \;\vee\; p. $$

So $$S_1$$ becomes

$$ (\sim p \;\vee\; q)\;\vee\;(q \;\vee\; p). $$

Using the associative and commutative laws of $$\vee,$$ we may regroup and list all literals together:

$$ (\sim p \;\vee\; p)\;\vee\;(q \;\vee\; q). $$

Now, by the law of excluded middle, $$\sim p \;\vee\; p = \text{T}$$ (always true).
Also, $$q \;\vee\; q = q.$$

Hence

$$ (\sim p \;\vee\; p)\;\vee\;(q \;\vee\; q)=\text{T}\;\vee\; q=\text{T}. $$

Because the disjunction reduces to the constant true value $$\text{T},$$ the formula is true for every possible truth assignment of $$p$$ and $$q.$$ Therefore $$S_1$$ is indeed a tautology.

Now we examine $$S_2 : (p \wedge \sim q)\;\wedge\;(\sim p \vee q).$$

For this conjunction to be true, both parts must be true simultaneously.

Start with the left conjunct $$p \wedge \sim q.$$ For it to be true we must have

$$p = \text{T} \quad\text{and}\quad q = \text{F}.$$

With these same truth values, evaluate the right conjunct $$\sim p \vee q.$$ Since $$p=\text{T},$$ we have $$\sim p=\text{F};$$ and with $$q=\text{F},$$ we obtain

$$\sim p \vee q = \text{F} \vee \text{F} = \text{F}.$$

Thus, whenever the first conjunct is true, the second conjunct is forced to be false, making the whole conjunction false. No other truth assignment can make the first conjunct true, so there is no assignment that makes the entire expression true.

Therefore the statement $$(p \wedge \sim q)\;\wedge\;(\sim p \vee q)$$ is always false; it is a fallacy. Hence the meta-statement $$S_2$$ saying “this formula is a fallacy” is true.

We have established that $$S_1$$ is a tautology (true) and $$S_2$$ correctly asserts a fallacy (so $$S_2$$ is also true). Therefore, both statements are true.

Hence, the correct answer is Option D.