There are lots of sources for homological algebra so it's hard to respond without knowing what you're reading. The independence of $\text{Ext}$ from the choice of resolution takes some elbow grease to show. See section 3 and 4 of these notes (or any text on homological algebra, Dummit and Foote, etc.).
The idea is that the identity map on $A$ can be extended to a map from one resolution to another. Writing $P_*$ and $Q_*$ for resolutions of $A$, we get maps from $P_*$ to $Q_*$ and $Q_*$ to $P_*$. One can show that the composition of these maps is chain homotopic to the identity map on $P_*$.
But the argument that homotopy equivalent complexes have isomorphic homologies is straightforward:
Suppose $(C, \partial)$ and $(D, \partial')$ are homotopy equivalent via chain maps $f: C \to D$ and $g: D \to C$. This means that $g \circ f - \text{Id}_C = \partial h + h \partial$ for some map $h: C_* \to C_{*+1}$. Now suppose that $\omega \in C$ is a cycle so that $[\omega] \in H(C)$ is a homology class. $$(\partial h + h \partial)_* [\omega] = [\partial h \omega + h \partial \omega] = [0]$$ because the homology class of a boundary is always $0$ and $\omega$ is assumed to be a cycle. So $(g\circ f - \text{Id}_C)_*[\omega] = 0$. But of course $(\text{Id}_C)_*$ is the identity map on homology, so $(g \circ f)_* = g_* \circ f_*$ is also the identity map. This implies that $f_*$ is injective and $g_*$ is surjective. Now apply this same argument to $f \circ g$ to see that $f_*$ and $g_*$ are isomorphisms. So $H_*(C) \simeq H_*(D)$.
(All of that is written with homology in mind -- I hope I got the gradings right -- but it applies mutatis mutandis to cohomology.)
