So i'm completely able to do the question below that relates to the picture in a few different ways. I know that $H_n(X,A) \cong \tilde{H}_n(X/A)$ since $X$ has a neighborhood that deformation retracts onto $A$, and so we get a wedge product which then breaks up into a direct product of homology groups, right.
Anyway, I also did the problem by looking at the long exact sequence for the pair (X,A) and I have a question...
Just looking at the part of the long exact sequence:
$... \rightarrow H_2(X) \rightarrow H_1(A) \rightarrow^a H_1(X) \rightarrow H_1(X,A) \rightarrow ...$
The map $a$ has image zero because $A$ bounds a subsurface of the space $X$ (a genus 2 oriented surface). However, $B$ does not bound such a subsurface, so in that case things are different. And this is what I need somebody to give me insight, what exactly is going on here? Why does $A$ bound a subsurface but $B$ doesn't? Thanks in advance!!
A closed $1$-manifold $C \subset X$ bounding a subsurface of $X$ means that there is a subsurface with boundary $S \subset X$ such that $C = \partial S$. Note that $\partial S$ agrees with the topological boundary $\text{bd}_XS$ of the subspace $S \subset X$
Note that neccesarily $U = X \setminus S$ is a nonempty open subset of $X$ (otherwise $\partial S = \partial X = \emptyset$). Hence $S' = U \cup \partial S = \text{cl}_X U$ is also a subsurface of $X$ which bounds $C$. We have $S \cap S' = C$. Thus $C$ separates $X$ in two submanifolds with boundary $S, S'$ having $C$ as their common boundary and whose interiors are disjoint. In particular, $X \setminus C$ is not connected.
Obviously $A$ bounds a subsurface of $X$ (in fact it bounds the left and the right half of $X$). What about $B$? $X \setminus B$ is connected, therefore it does not bound a subsurface.