Proceed methodically: Suppose the premisses are true and conclusion false. So
$1.\quad p \quad \Rightarrow \quad T$
$2.\quad p \lor q \quad \Rightarrow \quad T$
$3.\quad q \to (r \to s) \quad \Rightarrow \quad T$
$4.\quad t \to r \quad \Rightarrow \quad T$
$5.\quad \neg s \to \neg t \quad \Rightarrow \quad F$
From the last, you know
$6.\quad \neg s \quad \Rightarrow \quad T$
$7.\quad \neg t \quad \Rightarrow \quad F$
Whence
$8.\quad s \quad \Rightarrow \quad F$
$9.\quad t \quad \Rightarrow \quad T$
4 and 9 give us
$10.\quad r \quad \Rightarrow \quad T$
So $r \to s$ is false, and hence, from (3)
$11.\quad q \quad \Rightarrow \quad F$.
So we've worked backwards to successfully find a valuation of all the variables (at lines 1, 8--11) which you can check makes all of 1 to 5 true, i.e. makes the premisses true and conclusion false.
Systematizing this "working backwards" method gives us the user-friendly method of "semantic tableaux" or "truth-trees" used in many textbooks (including mine, and Paul Teller's which is freely available online).
I'll give a summary "sketch" of one proof approach.
I'd suggest you start from the first premise:
First check what follows from $r\land \lnot s$ and then what follows when $q \land \lnot s$. (At least one of these must be true. Why?)
From each of the above, we can use conjunction elimination (or simplification) to obtain $\lnot s$. And so, we can use disjunction elimination to conclude $\lnot s$.
From $\lnot s$ and premise $(2)$, we can derive (infer), by modus ponens, $(p\land r)\rightarrow u$.
From $(p \land r) \rightarrow u$ together with premise $(3)$, we get $(p\land r) \rightarrow (s \land \lnot t)$
Now, we already derived $\lnot s.$ This means $\lnot s \lor t$ is true. Why?
But $\lnot s \lor t \equiv \lnot (s \land \lnot t).$
From $\lnot (s \land \lnot t),$ together with our derived $(p\land r )\rightarrow (s\land \lnot t)$, by modus ponens, we get $\not(p \land r)$
This means that either $\lnot p$, or $\lnot r$ is true.
If $\lnot p$ is true, we can build from this (addition) to get $\lnot p \lor q \equiv p\rightarrow q$, as desired.
If $\lnot r$ is true, then $r\land \lnot s$ is false (first alternative from premise $(1)$. Then it must follow that $q\land \lnot s$ is true. So it follows that $q$ is true (simplification). And given $q$, we have $\lnot p \lor q$ (addition), which is equivalent to $p \rightarrow q$, as desired.
Therefore, from $\lnot (p \land r)\equiv \lnot p \lor \lnot r$ we can conclude $(q\rightarrow q)$.
I'll leave this to you to write formally (including keeping proper track of temporary assumptions), with proper justifications.
Best Answer
Rewrite the $p \lor r$ as $\neg \neg p \lor r$, which can then be rewritten as $\neg p \to r$
Also, $p \to q$ can be rewritten as $\neg q \to \neg p$
And now it is just a bunch of Hypothetical Syllogisms and you're pretty much there.
Formally:
$$ \begin{array}{lll} 1)&p\to q&\text{Premise}\\ 2)&r\to s&\text{Premise}\\ 3)&p\lor r&\text{Premise}\\ 4)&\neg \neg p\lor r&\text{Double Negation 3)}\\ 5)&\neg p\to r&\text{Conditional equivalence 4)}\\ 6)&\neg q\to \neg p&\text{Contraposition 1)}\\ 7)&\neg q\to r&\text{Hypothetical Syllogism 5,6)}\\ 8)&\neg q\to s&\text{Hypothetical Syllogism 2,7)}\\ 9)&\neg \neg q\lor s&\text{Conditional equivalence 8)}\\ 10)&q\lor s&\text{Double Negation 9)}\\ \end{array} $$