For the first proof, I actually finished but I'm not sure if it is correct.
Proof for $\neg q \rightarrow \neg p \vdash p \rightarrow q$:
$\neg q \rightarrow \neg p$, $[p]^1$, $[\neg q]^2$
$\neg p$ by $(\rightarrow E)$
$\bot$ by $(\bot I)$ obs:$(\neg p\ and\ p)$
$q$ by $(\rightarrow E)^2$ obs:(assuming ¬q and using the contradiction)
$p \rightarrow q$ by $(\rightarrow I)^1$
I have many issues with this proof because I used the assumptions $[p]^1$, $[\neg q]^2$ before using the rule that allows me to make assumptions (which in this case is $(\rightarrow I)^i$. I did that because I found it's easier to do the proof from bottom up, but it is not clear to me that this is allowed.
For the second proof, I haven't finished yet. I've been trying to prove it from bottom up, and I'm assuming that I should find a contradiction that would allow me to find ¬p and ¬q but I don't know if I'm in the right path.
There is no such thing as allowing to make assumptions. You can make assumptions any time you want. All you need to make sure is to eventually discharge any assumptions that you don't want occurring on the left-hand side of the $\vdash$. Your proof is correct.
Your start for the second proof is also correct. You're already halfway there. Hint: Place your premise $\neg (p \lor q)$ as an assumption. With this, what formula do you think you can get from $p$ to derive the desired contradiction?