$\text{Ext}^1(-,B)$, independence of projective resolution. If we want to compute the group $\text{Ext}^1(A,B)$ we take a projective resolution of $A$
$$\cdots\to P_2 \to P_1 \to P_0 \to A \to 0,$$ apply the contravariant functor $\text{Hom}(\cdot,B)$ to it (without $A$) and take 1st cohomology of the resulting chain complex. If we take another projective resolution then $\text{Ext}^1(A,B)$, even all $\text{Ext}^i$, remains isomorphic.
Why?  One argument resorts to saying that the two resolution are chain homotopic, but
I don't see why they are and how the isomorphism follows.  
As this is crucial part for the definition of Ext, I would like someone to shed some light on it.
 A: 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.)
A: P.J.Hilton, U.Stammbach, A Course in Homological Algebra: 
Corollary 3.5. If $\phi\cong\psi: C\to D$ and if $F$ is an additive functor, then 
$H(F\phi) = H(F\psi):H(FC)\to H(FD)$. 
