Let $A$ be an abelian category and $X$, $Y$ two objects of $A$. Let's define Ext in this way:
Ext$^i_A(X,Y)$=Hom$_{D(A)}(X[0],Y[i])$
Where $X[0]$ is the complex with all zeros except in degree 0 where it has $X$, and $Y[i]$ is the complex with all zeros except in degree $-i$ where it has $Y$.
Now I would like to prove that this definition is equivalent to the usual definition of Ext, i.e.:
take a projective resolution of $X$: $\cdots\rightarrow P^{-1}\rightarrow P^0\rightarrow X\rightarrow 0$ then Ext$^n_A(X,Y)$ is the $n$th cohomology group of the complex $0\rightarrow\mathrm{Hom}(P^0,Y)\rightarrow\mathrm{Hom}(P^{-1},Y)\rightarrow\mathrm{Hom}(P^{-2},Y)\rightarrow\cdots$.
I have an hint: denote by $K(A)$ the homotopy category. It seems to be useful to prove that we have an isomorphism
$\mathrm{Hom}_{K(A)}(X^\bullet,Y^\bullet)\rightarrow\mathrm{Hom}_{D(A)}(X^\bullet,Y^\bullet)$ in the following cases:
1) $Y^\bullet\in\mathrm{Ob}\;Kom^+(I)$, i.e. $Y^\bullet$ is a bounded complex on the left of injectives objects;
2) $X^\bullet\in\mathrm{Ob}\;Kom^-(P)$, i.e. $X^\bullet$ is a bounded complex on the right of projective objects.
I'm pretty sure that we need only one between 1 and 2, and the other is useful if we want to prove the caracterization of Ext with injective resolutions, but I can do that if you could show me how to do it with projective resolutions.
By the way the map $\mathrm{Hom}_{K(A)}(X^\bullet,Y^\bullet)\rightarrow\mathrm{Hom}_{D(A)}(X^\bullet,Y^\bullet)$ is $f\mapsto$ the equivalence class of the roof $X\leftarrow X\rightarrow Y$, where $id:X\rightarrow X$ and $f:X\rightarrow Y$.
And if you need the definition of roof just look to one of my previous questions: why is this composition well defined?
Best Answer
$\DeclareMathOperator{Hom}{Hom}$The main point is that a projective resolution $\dots \to P^{-2} \to P^{-1} \to P^{0} \xrightarrow{\alpha^0} X$ gives you a quasi-isomorphism $\alpha \colon P^{\bullet} \to X[0]$. Indeed, the mapping cone of $\alpha$ is exact because it is the resolution (maybe up to an immaterial sign in the differentials).
A quasi-isomorphism becomes an isomorphism in the derived category (because that's what we invert), in particular precomposition with $\alpha^{-1}$ (or, if you prefer: composition with the roof $X[0] \xleftarrow{\alpha} P^\bullet \xrightarrow{1} P^\bullet$) gives an isomorphism $$\Hom\nolimits_{D(A)}(P^\bullet,Y) \xrightarrow{\cong} \Hom\nolimits_{D(A)}(X[0],Y) $$ for every complex $Y \in D(A)$.
From fact 2) of the question we have a composite isomorphism $$ \Hom\nolimits_{K(A)}(P^\bullet,Y) \xrightarrow{\cong} \Hom\nolimits_{D(A)}(P^\bullet,Y) \xrightarrow{\cong} \Hom\nolimits_{D(A)}(X[0],Y) $$ for all complexes $Y$ which is explicitly given by sending a (homotopy class of a) chain map $f\colon P^\bullet \to Y$ to the roof $X[0] \xleftarrow{\alpha} P^\bullet \xrightarrow{f} Y$, so it only remains to identify the complex $\Hom_{K(\mathscr{A})}(P^\bullet, Y)$.
I hope I got the signs and indices right in what is to follow: for any two complexes $A$ and $B$ over an additive category $\mathscr{A}$ one defines the total $\Hom$-complex $\Hom\nolimits^\bullet(A,B)$ of abelian groups by $$ \Hom\nolimits^k(A,B) = \prod_{n \in \mathbb{Z}}\Hom\nolimits_{\mathscr{A}}(A^n,B^{n+k}) $$ with differential $(f^n)_{n \in \mathbb{Z}} \mapsto \left(f^{n+1}d_A^n-(-1)^{k}d_B^{n+k}f^n \right)_{n\in\mathbb{Z}}$ and it is elementary to check (if indeed I got the signs right) that $$\boxed{ H^k\left(\Hom\nolimits^\bullet(A,B)\right) = \Hom\nolimits_{K(\mathscr{A})}(A,B[k]) }$$ because in order for $f = (f_n)_{n \in \mathbb{Z}}$ to be a cycle in $\Hom\nolimits^k(A,B)$ it is necessary and sufficient that $f \colon A \to B[k]$ defines a chain map and to be a boundary means that it is homotopic to zero. (In the sign conventions I'm used to shifting a complex by $1$ involves multiplying its differentials by $-1$)
If we take $A = P^\bullet$ and $B=Y[0]$ and $k\in\mathbb{Z}$ we see that the $\Hom$ complex collapses to $\Hom^k(P^{\bullet},Y[0]) = \Hom\nolimits_{\mathscr{A}}(P^{-k},Y)$ and computing its cohomology amounts to taking the cohomology of the complex $$ \dots \to 0 \to \Hom\nolimits_{\mathscr{A}}(P^0,Y) \xrightarrow{d^\ast} \Hom\nolimits_{\mathscr{A}}(P^{-1},Y) \xrightarrow{d^\ast} \Hom\nolimits_{\mathscr{A}}(P^{-2},Y) \xrightarrow{d^\ast} \cdots $$ which via $$ \operatorname{Ext}^k(X,Y) = \operatorname{Hom}_{D(A)}(X,Y[k]) \cong \operatorname{Hom}_{K(A)}(P^\bullet,Y[k]) = H^k(\operatorname{Hom}^\bullet(P^\bullet,Y[0])) $$ gives the identification you ask about.