Suppose $p$, $q$ and $r$ are propositions such that $p \to q$, $(\neg p) \to r$ and $r \to (p \lor q)$ are all true. Show that $q$ is true 
I don't understand how this shows $q$ is true.
I understand we assume $q$ is false. Then in order for $p\to q$ to be true, $p$ must also be false.
Then, in order for $(\lnot p) \to r $ to be true, $r$ must be true because $p$ is already false, so negating it would have it be true; thus $r$ must be true.
But I don't understand the last one.
$ r \to (p \lor q)$, we know $r$ is true, but what about $p \lor q$?
 A: What about $p\lor q$?
Since $q$ is false (by assumption) and $p$ is false (since $p\implies q$ and $q$ is false), $p\lor q$ is false.
And if $r$ is true (as you acknowledged) but $p\lor q$ is false, then $r\implies(p\lor q)$ is false.
A: If $p$ is not true, so $r$ is true. So, $q$ is true, because either   $p$ or $q$ ($p\lor q$) is true when $r$ is.
A: First, assume $p$ is true.  Than $p \rightarrow q$ together with $p$ true forces $q$ to be true.
Now assume $p$ is false so that $\lnot p$ is true.  Then $\lnot p \rightarrow r$ forces $r$ to be true, which in turn (via $r \rightarrow (p \lor q)$) forces $p \lor q$ to be true.  But in this branch of the analysis we're assuming that $p$ is false.  If $p$ is false and $p \lor q$ is true, then $q$ must be true.
Therefore, whether $p$ is true or false, $q$ must be true.
A: 
I understand we assume $q$ is false. Then in order for $p\to q$ to be true, $p$ must also be false.
Then, in order for $(\lnot p) \to r $ to be true, $r$ must be true because $p$ is already false, so negating it would have it be true; thus $r$ must be true.
But I don't understand the last one.
$ r \to (p \lor q)$, we know $r$ is true, but what about $p \lor q$?

You already claimed to know $p$ is false under the assumption of $\neg q$, so you may conclude by disjunctive syllogism that $q$ is true in that context ; which contradicts the assumption.
Therefore the assumption is falsified, and so $\neg\neg q$ must be true, and hence...
A Fitch representation of what you have done:
$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\fitch{~~1.~p\to q
\\~~2.~\neg p\to r
\\~~3.~r\to (p\vee q)}{\fitch{~~4.~\neg q}{\fitch{~~5.~p}{~~6.~q\quad{\to}\mathsf E~5,1\\~~7.~\bot\quad\neg\mathsf E~6,4}\\~~8.~\neg p\qquad\neg\mathsf I~5{-}7\\~~9.~r\qquad~~~{\to}\mathsf E~2,9\\10.~p\vee q\quad{\to}\mathsf E~3,4\qquad\longleftarrow\text{You are here}\\\fitch{11.~q}{12.~\bot\quad{\neg}\mathsf E~11,4}\\13.~\bot\qquad~~{\vee}\mathsf E~10,5{-}7,11{-}12}\\14.~\neg\neg q\qquad~~~\neg\mathsf I~4{-}13\\15.~q\qquad\qquad\neg\neg\mathsf E~14
}$
A: Let me augment the proof.
Suppose $q$ is false. Then, by $p\to q$ and modus tollens, $p$ also has to be false. This will force $r$ to be true by $(\lnot p)\to r$. Also $p\lor q$ is false. Now, since a false statement cannot be implied by a true statement, the third formula $r\to(p\lor q)$ will be false. This is a contradiction. Hence $q$ is true.
A: 
$ r \to (p \lor q)$, we know $r$ is true, but what about $p \lor q$?

Remember that you were working under the assumption of $q$ being false, and you already showed that that means that $p$ is false as well. And, if $p$ and $q$ are both false, then $p \lor q$ is false as well. 
So, with $r$ being true, $r \to (p \lor q)$ becomes false.
Thus, the assumption that $q$ is false contradicts the last premise, and hence $q$ cannot be false.
