$\mathcal{Ext}^i(\mathcal{O}_{L_1}, \mathcal{O}_{L_2})$ and $\text{Ext}^i(\mathcal{O}_{L_1}, \mathcal{O}_{L_2})$ for two lines in $\mathbb{P}^3$ Let $L_1$ and $L_2$ be two lines in $\mathbb{P}^3$. I want to compute $\mathcal{Ext}^i(\mathcal{O}_{L_1}, \mathcal{O}_{L_2})$ and $\text{Ext}^i(\mathcal{O}_{L_1}, \mathcal{O}_{L_2})$. Basically, there are three cases: $L_1$ and $L_2$ coincide, $L_1$ and $L_2$ intersect and $L_1$ and $L_2$ do not intersect.
For convenience suppose that $L_1=\{z_2=z_3=0\}$, where $z_0,z_1,z_2,z_3$ are the homogeneous coordinates. The we have the following locally-free resolution for $\mathcal{O}_{L_1}$:
$$0\to\mathcal{O}_{\mathbb{P}^3}(-2)\stackrel{\left(
\begin{array}{c}
z_2\\
-z_3\\
\end{array}
\right)}\longrightarrow\mathcal{O}_{\mathbb{P}^3}(-1)\oplus\mathcal{O}_{\mathbb{P}^3}(-1)\stackrel{(z_3\,z_2)}\longrightarrow\mathcal{O}_{\mathbb{P}^3}\to\mathcal{O}_{L_1}\to0. \,\,\,(1)$$
In order to compute $\mathcal{Ext}^i(\mathcal{O}_{L_1}, \mathcal{O}_{L_2})$ I need to apply $\mathcal{Hom}(\_, \mathcal{O}_{L_2})$ to (1) and to compute cohomologies of the complex
$$0\to\mathcal{O}_{L_2}\stackrel{\left(
\begin{array}{c}
z_3\\
z_2\\
\end{array}
\right)}\longrightarrow\mathcal{O}_{L_2}(1)\oplus\mathcal{O}_{L_2}(1)\stackrel{(z_2\,-z_3)}\longrightarrow\mathcal{O}_{L_2}(2)\to0.\,\,\,(2)$$
Now if $L_2=L_1$ for example, then it seems that all maps in (2) are equal to 0 and then we have $\mathcal{Ext}^0(\mathcal{O}_{L_1}, \mathcal{O}_{L_1})=\mathcal{O}_{L_1}$, $\mathcal{Ext}^1(\mathcal{O}_{L_1}, \mathcal{O}_{L_1})=\mathcal{O}_{L_1}(1)\oplus\mathcal{O}_{L_1}(1)$, $\mathcal{Ext}^2(\mathcal{O}_{L_1}, \mathcal{O}_{L_1})=\mathcal{O}_{L_1}(2).$
Is it correct?
Suppose now that $L_1$ and $L_2$ do not intersect. Then we can assume that $L_2=\{z_0=z_1=0\}$. 
I have some difficulties computing cohomologies of (2).
Also, how could I compute $\text{Ext}^i(\mathcal{O}_{L_1}, \mathcal{O}_{L_2})$?
$\textbf{Update}$
It seems that I need to use local-to-global $\text{Ext}$ spectral sequence
$$H^p(X, \mathcal{Ext}^q(F,G))\Rightarrow\text{Ext}^{p+q}(F,G).$$
Can anyone help me with this?
 A: To compute the $\operatorname{Ext}$ in this case, you can indeed use the spectral sequence
$$H^p(X, \mathscr Ext^q(\mathscr F, \mathscr G)) \Rightarrow \operatorname{Ext}^{p+q}(\mathscr F, \mathscr G),$$
arising as the Grothendieck spectral sequence of the composition of functors
$$\operatorname{Mod}_{\mathcal O_X} \to \operatorname{Mod}_{\mathcal O_X} \to \operatorname{Ab},$$
where the first functor is $\mathscr Hom_{\mathcal O_X}(\mathscr F, -)$, and the second is $\Gamma$. (We will drop the $\mathcal O_X$ from the subscript in $\mathscr Hom$ and $\operatorname{Hom}$ when $X$ is understood.)
Remark. (The French would put a 'dangerous bend' road sign here.) You cannot in general define $\mathscr Ext$ as derived functors of $\mathscr Hom$ on $\operatorname{Qcoh}(X)$ or $\operatorname{Coh}(X)$. Indeed, $\mathscr Hom(\mathscr F, \mathscr G)$ is not in general quasicoherent if $\mathscr F$ and $\mathscr G$ are. However, it is if $\mathscr F$ is coherent (Hartshorne, Exercise III.6.3(b)), and you could define $\mathscr Ext$ on $\operatorname{Qcoh}(X)$. You then need to check that it gives the same result, which involves showing that the above definition gives a coerasable $\delta$-functor on $\operatorname{Qcoh}(X)$. You can use the construction from Hartshorne Corollary III.3.6 and the computation from Proposition III.6.8 to do this. (All of this uses a Noetherian assumption, which is fine for our purposes.)
However, I propose just sticking with the above definition, which I think is standard.
Remark. In order to have a Grothendieck spectral sequence, we have to show that $\mathscr Hom$ takes injectives to $\Gamma$-acyclic sheaves. Godement gives the following argument (in his Topologie algébrique et théorie des faisceaux): if $\mathscr I$ is an injective $\mathcal O_X$-module, then for any $\mathcal O_X$ module $\mathscr F$, and any open $U \subseteq X$, we have an injection
$$\mathscr F_U \to \mathscr F,$$
where $\mathscr F_U = j_!(\mathscr F|_U)$, for $j \colon U \to X$ the inclusion. Thus, as $\mathscr I$ is injective, any morphism $\mathscr F_U \to \mathscr I$ extends to a morphism $\mathscr F \to \mathscr I$. Show that this implies that
$$\operatorname{Hom}_{\mathcal O_X}(\mathscr F, \mathscr I) \to \operatorname{Hom}_{\mathcal O_U}(\mathscr F|_U, \mathscr I|_U)$$
is surjective, i.e. $\mathscr Hom (\mathscr F, \mathscr I)$ is flasque.
Example. Let's compute the spectral sequence for $\mathscr Ext^i(\mathcal O_{L_1}, \mathcal O_{L_1})$. I claim that it collapses on the $E_2$ page (i.e. there is only one nonzero row or column):
\begin{align*}
H^p(X, \mathscr Ext^q(\mathcal O_{L_1}, \mathcal O_{L_1})) &= H^p(X, \bigwedge\nolimits^q\mathcal O_{L_1}(q))\\
&= H^p(L_1, \bigwedge\nolimits^q\mathcal O_{L_1}(q)),
\end{align*}
which is zero is $p > 0$, as $H^1(\mathbb P^1, \mathcal O(d)) = 0$ for $d \geq 0$ (in fact, for $d \geq -1$). Thus,
$$\operatorname{Ext}^q(\mathcal O_{L_1}, \mathcal O_{L_1}) = \Gamma(L_1, \bigwedge\nolimits^q\mathcal O_{L_1}(q)).$$
It's easy to write down explicit bases for these.
Example. To compute $\operatorname{Ext}(\mathcal O_{L_1},\mathcal O_{L_2})$ when $L_1$ and $L_2$ do not intersect, note that then sequence (2) is exact, since $z_2, z_3$ forms a regular sequence in $S = \Gamma_*(L_2, \mathcal O_{L_2})$, and the scheme they cut out is empty (as $L_1 \cap L_2 = \varnothing$). Thus, the sheaf Exts vanish, hence by the spectral sequence so do the global Exts.
Example. Finally, if $L_1$ and $L_2$ intersect, they do so in a point $P$, and $L_1$ and $L_2$ are coplanar. We may assume that $L_2$ is now given by $z_1 = z_2 = 0$. One easily computes that
$$\mathscr Ext^q(\mathcal O_{L_1},\mathcal O_{L_2}) = \left\{\begin{array}{ll}0, & q = 0,\\ \mathcal O_P(1), & q = 1,\\ \mathcal O_P(2), & q = 2.\end{array}\right.$$
But on a point $P$, the Serre twists don't matter. Now the Ext spectral sequence trivially collapses on the $E_2$ page, because a point has no higher cohomology. Thus,
$$\dim_k \operatorname{Ext}^q(\mathcal O_{L_1},\mathcal O_{L_2}) = \left\{\begin{array}{ll}0, & q = 0,\\ 1, & q = 1,2. \end{array}\right.$$
Edit: Indeed, the first map in (2) is injective, showing the result I claim for $q = 0$. The kernel of the second map is the first copy of $\mathcal O_{L_2}$, whereas the image of the first map is the things in there that are divisible by $z_3$. Finally, the image of the second map is again the things that are divisible by $z_3$.
