If the truth value of the compound statement $[(p \leftrightarrow q) \wedge (q \to r) \wedge \sim r] \to (p \wedge \sim q)$ is false, then the truth values of the statement patterns $(p \to q) \leftrightarrow (q \to r)$ and $\sim(p \vee r) \to (q \wedge p)$ are, respectively ...

The given compound statement is $[(p \leftrightarrow q) \wedge (q \to r) \wedge \sim r] \to (p \wedge \sim q)$.

An implication $X \to Y$ is false if and only if $X$ is true and $Y$ is false.

Thus, $(p \leftrightarrow q) \wedge (q \to r) \wedge \sim r$ is true and $p \wedge \sim q$ is false.

For the conjunction $(p \leftrightarrow q) \wedge (q \to r) \wedge \sim r$ to be true, each of its components must be true:
$\sim r$ is true $\Rightarrow r = F$
$q \to r$ is true $\Rightarrow q \to F$ is true $\Rightarrow q = F$
$p \leftrightarrow q$ is true $\Rightarrow p \leftrightarrow F$ is true $\Rightarrow p = F$

Checking the second part with these values: $p \wedge \sim q = F \wedge T = F$, which satisfies the condition that $Y$ is false.

So, the truth values are $p = F$, $q = F$, $r = F$.

Now, evaluating the first statement pattern:
$(p \to q) \leftrightarrow (q \to r)$
$= (F \to F) \leftrightarrow (F \to F)$
$= T \leftrightarrow T = T$

Evaluating the second statement pattern:
$\sim(p \vee r) \to (q \wedge p)$
$= \sim(F \vee F) \to (F \wedge F)$
$= \sim F \to F$
$= T \to F = F$

The truth values are $T$ and $F$, respectively.

Answer: $(T, F)$