Let $X$ be a topological space and $A\subset X$ closed as well as $A \subset U \subset X$ with $U$ open. Consider the quotient map $q:X\rightarrow X/A$. My question is that I don't know how to prove that
$$q|_{X\smallsetminus A}:X \smallsetminus A\rightarrow (X/A) \smallsetminus (A/A)$$
and
$$q|_{U\smallsetminus A}:U \smallsetminus A\rightarrow (U/A) \smallsetminus (A/A)$$
are homeomorphisms. Showing that they are bijective is quite trivial, but I don't really know how to prove that they are homeomorphisms. I definitely have to use that $A$ is closed and $U$ is open as this statement is not true otherwise. Thanks for helping!
If $B\subseteq X$ is any subset then
$$q^{-1}(q(B))=\begin{cases} B\cup A&\text{if }B\cap A\neq\emptyset \\ B&\text{otherwise} \end{cases}$$
Therefore if $B\subseteq X\backslash A$ then $q^{-1}(q(B))=B$. Thus if additionally $B$ is open in $X$ then $q(B)$ is open in $X/A$. Next since $A$ is closed, then $B$ is open in $X\backslash A$ if and only if it is open in $X$. And therefore $q_{|X\backslash A}$ is an open map. It is of course continuous, and therefore a homeomorphism (since you've correctly noted it is a bijection).
As for the second question, lets first note the subtlety here. Our $U/A$ can be given two topologies: first, the subspace topology inherited from $X/A$, and second: the quotient topology from $U/A$. And normally an answer would depend on that. However we have a special situation here.
If we consider the quotient topology, then the answer is quite trivial: it is literally the same reasoning as before, you just replace $X$ with $U$.
On the other hand with subspace topology, the answer is only slightly harder. Normally these two don't have to coincide, but they do sometimes. A consequence of Theorem 22.1 from Munkres' "Topology" is that this happens whenever $q$ is closed and $U$ is saturated.
So first of all $q$ is closed, because given $F\subseteq X$ we have $q^{-1}(q(F))$ is either $F$ or $F\cup A$, hence closed. And thus $q(F)$ is closed.
Finally $U$ is saturated. And that's because $q^{-1}(q(U))=U\cup A=U$. Both equalities because $A\subseteq U$.
Note that it doesn't matter that $U$ is open, this holds for any $U$ as long as $A\subseteq U$.