Where we use assumption that resolution is locally free (Hartshorne III.6.5) Where we use assumption that resolution of $\mathcal{G}$ must be locally free of finite rank? It seems natural to me, that $h^0(\mathcal{Hom(L.,G))}$ must be equal to $\mathcal{Hom(F,G)}$ without that assumption (thats because $\mathcal{Hom(?,G)}$ is left exact contravariant functor), and $\mathcal{Hom(?,I)}$ must be exact whenever $\mathcal{I}$ is injective object without that assumption too (thats just another definition of injective object). I am confused, hope for your help!

 A: If $\mathscr{L}_.$ is not a complex of locally free sheaves, then $h^i(\mathscr{H}om(\mathscr{L}_.,\mathscr{G}))$ is not a $\delta$-functor in $\mathscr{G}$. Indeed, given a short exact sequence $0\rightarrow\mathscr{G}'\rightarrow\mathscr{G}\rightarrow\mathscr{G}''\rightarrow 0$, the sequence
$$0\rightarrow\mathscr{H}om(\mathscr{L}_i,\mathscr{G}')\rightarrow\mathscr{H}om(\mathscr{L}_i,\mathscr{G})\rightarrow\mathscr{H}om(\mathscr{L}_i,\mathscr{G}'')\rightarrow 0$$
must be exact for every $i$. Otherwise, there won't be any connecting homomorphism.
If the $\mathscr{L}_i$ are locally free, then $\mathscr{H}om(\mathscr{L}_i,.)$ are exact functors, so the right hand side is indeed a $\delta$-functor.
By the way, if you take the trivial resolution of $\mathscr{F}$, namely $0\rightarrow\mathscr{F}\rightarrow\mathscr{F}\rightarrow 0$, then $h^i(\mathscr{H}om(\mathscr{L}_.,\mathscr{G}))$ would be $0$ for $i>0$. Clearly this is not right, it can't be isomorphic to $\mathscr{E}xt^i(\mathscr{F,G})$ otherwise $\mathscr{E}xt^i$ would not be interesting at all.
A: I was wondering about this only this morning! Could somebody be so kind as to read my interpretation and tell me if it is correct or not?
First, $\mathcal{Hom}(\mathcal L, \ . \ )$ is an exact functor if $\mathcal L$ is locally free. This is because $\mathcal{Hom}(\mathcal L, \ . \ )$ locally looks the same as $\mathcal{Hom}(\mathcal O, \ . \ )$ - just restrict to a trivialising neighbourhood.
So given a short exact sequence,
$$ 0 \to \mathcal F \to \mathcal G \to \mathcal H \to 0,$$
we get a short exact sequence of complexes,
$$ 0 \to \mathcal{Hom} (\mathcal L^\bullet, \mathcal F) \to \mathcal{Hom} (\mathcal L^\bullet, \mathcal G)\to \mathcal{Hom} (\mathcal L^\bullet, \mathcal H) \to 0.$$
This short exact sequence induces a long exact sequence by the snake lemma (which I believe is valid in all abelian categories):
$$ \dots \to h^i (\mathcal{Hom} (\mathcal L^\bullet, \mathcal F) ) \to h^i (\mathcal{Hom} (\mathcal L^\bullet, \mathcal G) ) \to h^i ( \mathcal{Hom} (\mathcal L^\bullet, \mathcal H) )  \to h^{i+1} (\mathcal{Hom} (\mathcal L^\bullet, \mathcal F) )  \to \dots $$
I believe the existence of this long exact sequence is what Hartshorne is refering to when he says that $h^i(\mathcal {Hom}, . \ )$ is a $\delta$-functor.
Like I said, I would be very grateful indeed if somebody could check this - even if there is a better answer to the question.
