$$\displaylines{ ¬p → (r ∧ ¬s)\cr t → s\cr u → ¬p\cr ¬w\cr u ∨ w\cr ∴ t → w\cr}$$
I have the solution which shows
We start by noticing that we have (How did we know we have to start here?) $$\displaylines{ u ∨ w; ¬w; ∴ u.\cr}$$
Indeed, if u ∨ w and ¬w are both true, then w is false, and u must be true.
Next $$\displaylines{ u → ¬p; u; ∴ ¬p.\cr}$$
Indeed, if u → ¬p is true, either u is true and ¬p is true, or u is false. But u is true, thus ¬p is true (Modus Ponens).
Then $$\displaylines{ ¬p → r ∧ ¬s; ¬p; ∴ r ∧ ¬s,\cr}$$
this is again Modus Ponens. Then $$\displaylines{r ∧ ¬s; ∴ ¬s.\cr}$$
Indeed, for r ∧ ¬s to be true, it must be that ¬s is true. Finally [Why can't let r to be false instead?)
$$\displaylines{t → s; ¬s; ∴ ¬t\cr}$$
since for t → s to be true, we need either t to false, or t and s to be true, but since s is false, t must be false (Modus Tollens), and
$$\displaylines{¬t ∴ ¬t ∨ w\cr}$$ or equivalently $$\displaylines{¬t ∨ w ≡ t → w\cr}$$ using the Conversion theorem, which shows that the argument is valid.
There is really no particular way you can know this, these problems generally involve a bit of trial and error.
$\wedge$ means "and": if $r\wedge\neg s$ is true then $r$ is true and $\neg s$ is true.
But anyway I think this is a very complicated solution. Better: mark the statements (1), (2) etc, then
That is, assuming $t$ true shows $w$ true; so the hypothetical $t\to w$ is true.
In fact, one can say more: assuming $t$ is true we have $w$ true as above; but (4) says $w$ is false which is a contradiction. So in fact $t$ is false, and this means that $t\to\langle\hbox{anything}\rangle$ is a valid conclusion!