How do you approach solving problems like these using rules of inference? So using rules of inference can you show that the premises:
$$\neg p\vee q\rightarrow r,\;s\rightarrow\neg q,\;p\rightarrow t,\;\neg t,\;\neg p\wedge r\rightarrow s$$
lead to the conclusion $\neg q$?
How would one approach trying to solve a problem like this? Mainly how do you know which premise to start with and the combinations to use to get to the conclusion? Or is it just one of those things that require time and practice to get the hang of. 
 A: There are only so many rules you can apply at a time.  Without anything clever (like using cases), the only rule of inference you can apply immediately is modus tollens $p\to t, \neg t$.  This gives you $\neg p$, and now you can apply it to the first implication to get $r$.  Now you have $\neg p$ and $r$, so using the last implication you get $s$.  Now using $s$ and the second implication, you get $\neg q$.
A: Let's first list the premises with some tags to refer to them later:
\begin{align*}
\neg p \lor q \to r \tag{i} \\
s \to \neg q \tag{ii}\\
p \to t \tag{iii}\\
\neg t \tag{iv}\\
\neg p \land r \to s \tag{v}
\end{align*}
Note that the only apparent starting point is (iv).
Now (iv), with (iii), gives $\neg p$. This result and (i) implies $r$, and hence by (v) and these facts we have $s$. Finally we apply (ii) to obtain $\neg q$.
Alternatively, we could have worked backwards. Since (ii) is the only apparent ending point, we see that we only need to show $s$. From (v), it suffices to obtain $\neg p$ and $r$. Then (i) implies that we only need $\neg p$. From here we use (iii) and (iv) to obtain the result.
There is certainly a lot of trial and error, along with practice and experience, but you will usually want to use some combination of the above techniques as a sort of "general strategy". In other words, start with what you know and work a little. Start with what you want and work backwards a bit. Hopefully the two will meet in the middle somewhere.
A: I find writing them in a column instead of a row helpful.  Here it would make it easier to see that you have both $ p \implies t$ and $\lnot t$, while this tends to get lost when they are in a line.  It does take time and practice to see it easily.  Once you get $\lnot p$ from the first thing, it takes practice to see that you can get $r$ from the first but don't learn anything (yet) about $s$ from the last.  Now that you have $r$ you should see you can get $s$ from the last.  You need to be very comfortable with the symbols:  $\vee$ meaning or and $\wedge$ meaning and needs to be automatic.
A: 
¬p ∨ q → r, s → ¬q, p → t, ¬t, ¬p ∧ r → s;
lead to the conclusion ¬q?

Work backwards.   To lead to $\neg q$ we see that we have $s\to\neg q$ and so we need to get $s$.    To lead to $s$ we ...
$$\dfrac{\dfrac{\dfrac{\dfrac{p\to t\quad\neg t}{\phantom{\neg p}}{\tiny MT}\quad\dfrac{\dfrac{\dfrac{p\to t\quad\neg t}{\phantom{\neg p}}{\tiny MT}}{\phantom{\neg p\vee q}}{\tiny DI}\quad\lower{1.5ex}{\neg p\vee q\to r}}{\phantom r}{\tiny MP} }{\phantom{\neg p\wedge r}}{\tiny CI}\quad \lower{1.5ex}{\neg p\wedge r\to s}}s{\tiny MP} \quad \lower{1.5ex}{s\to \neg q}}{\neg q}{\tiny MP}$$
A: It depends on the rules you have, and even then it is a matter of practice: you need to learn general strategies, specific patterns, and other heuristics and tricks.
Doing formal proofs is like playing chess: like the rules of chess, the inference rules of the logic system you are using tell you what you can do, but not what you ought to do, or even what a useful to do would be. Like chess, you need practice and exposure to lots of different proofs to get good at that. Indeed, in chess you'll learn opening games, end games, general tactics, and specific combinations of moves ... and so it goes with proofs.
For example, with a conclusion of $\neg q$, my immediate reaction is: Proof by Contradiction!  That is assume $q$, and see if you can get a contradiction from that. Now, let's see ... you can combine $q$ and the first premise to quickly get to $r$ ... and I could combine that with the last premise ... if i also had $\neg p$ ... ok, so can I get to $\neg p$? Yes, that's just a modus tollens on $p \to t$ and $\neg t$. Great, so I can get $s$, and combining that with $s \to \neg q$ I get $\neg q$, which contradicts my assumption $q$. So yes, $q$ leads to a contradiction, and therefore we have proven $\neg q$
Now, was this the fastest way to prove this? No, vadim's answer shows a proof that does not use the proof by contradiction strategy, and is a little faster. However, that is a trade-off you'll also have to make: some strategies are pretty robust and reliable, but not always the most efficient.
