Vanishing of Ext groups of Coherent sheaves over Noetherian regular scheme Let $(X,\mathcal O_X)$ be a Noetherian regular scheme of dimension $1$.
Then, for any coherent sheaf $\mathcal F$ and any quasi-coherent sheaf $\mathcal G$, it holds that  $\mathcal Ext^i(\mathcal F, \mathcal G)=0, \forall i\ge 2$. Indeed, it is enough to show that all the stalks are zero, and now remembering $\mathcal Ext^i(\mathcal F, \mathcal G)_x \cong \text{ Ext}^i_{\mathcal O_{_x}}(\mathcal F_x, \mathcal G_x)$ and that each $\mathcal F_x$ is a module over the regular local ring $\mathcal O_x$ of dimension $\le 1$ so $\mathcal F_x$ has projective dimension at most $1$ over the regular local rings, we get the necessary vanishing.
Now my question is:

When can we say that the Ext groups $\mathrm{Ext}^i(\mathcal F,\mathcal G)=0$ for all $i\ge 2$ and for all coherent sheaves $\mathcal F,\mathcal G$ ?

Further thoughts: we also have the spectral sequence $$ E^{p,q}_2 = H^p(X, \mathcal{Ext}^q(\mathcal{F}, \mathcal{G})) \implies \mathrm{Ext}^{p+q}(\mathcal{F}, \mathcal{G}) $$ and so remembering sheaf cohomologies vanish after dimension step, one gets $E^{p,q}_2=0$ for $p+q>2$, hence $\mathrm{Ext}^i(\mathcal F,\mathcal G)=0, \forall i>2$. So one only needs to understand when $\mathrm{Ext}^2(\mathcal F,\mathcal G)$ vanish.
 A: You can use the local-to-global Ext spectral sequence,
$$ E^{p,q}_2 = H^p(X, \mathcal{Ext}^q_{\mathcal{O}}(\mathcal{F}, \mathcal{G})) \implies \mathrm{Ext}^{p+q}_{\mathcal{O}}(\mathcal{F}, \mathcal{G}) $$
Because $X$ has dimension $1$, $E^{p,q}_2 = 0$ for $p > 1$ and by regularity as you said $E^{p,q}_2 = 0$ for $q > 1$. Therefore, $E^{p,q}_2 = 0$ for $p + q > 2$ and therefore the convergence spectral sequence implies that $\mathrm{Ext}^{n}_{\mathcal{O}}(\mathcal{F}, \mathcal{G}) = 0$ for $n > 2$. However, if $\mathcal{F}$ is locally free (more generally flat) then $\mathcal{Ext}^q_{\mathcal{O}}(\mathcal{F}, \mathcal{G}) = 0$ for $q > 0$ and therefore the spectral sequence is converged giving,
$$ \mathrm{Ext}^{p}_{\mathcal{O}}(\mathcal{F}, \mathcal{G}) = H^p(\mathcal{Hom}_{\mathcal{O}}(\mathcal{F}, \mathcal{G})) = H^p(\mathcal{F}^\vee \otimes \mathcal{G}) $$
and therefore vanishes for $p > 1$.
Now, I claim that $\mathcal{Ext}^1_{\mathcal{O}}(\mathcal{F}, \mathcal{G})$ is supported at finitely many points. Indeed, for any coherent sheaf $\mathcal{F}$ there is a dense open $U$ such that $\mathcal{F}|_U$ is locally free. Therefore, $\mathcal{Ext}^1(\mathcal{F}, \mathcal{G})|_U = 0$ so $\mathcal{Ext}^1(\mathcal{F}, \mathcal{G})$ is supported on the complement $X \setminus U$ which is a finite set of points because $X$ is one dimensional noetherian and $U$ is dense and thus contains the generic point of each irreducible component (including every positive dimensional component).
Therefore, $H^1(X, \mathcal{Ext}^1(\mathcal{F}, \mathcal{G})) = 0$ because this sheaf is supported on a zero dimensional locus so from the spectral sequence we conclude that $\mathrm{Ext}^2_{\mathcal{O}}(\mathcal{F}, \mathcal{G}) = 0$.
EDIT: the following has a hole in it:
Now consider the case that $X$ is separated and connected. Then we can apply Hartshorne exercise III.6.8 to conclude that there are enough locally-free sheaves so for any coherent $\mathcal{O}_X$-module $\mathcal{F}$, there is a locally free resolution,
$$ 0 \to \mathcal{E}_1 \to \mathcal{E}_0 \to \mathcal{F} \to 0 $$
Then from the long exact sequence,
$$ \mathrm{Ext}^2_{\mathcal{O}}(\mathcal{F}, \mathcal{G}) \cong \mathrm{coker}{(\mathrm{Ext}^1_{\mathcal{O}}(\mathcal{E}_0, \mathcal{G}) \to \mathrm{Ext}^1_{\mathcal{O}}(\mathcal{E}_1, \mathcal{G}))} = \mathrm{coker}(H^1(X, \mathcal{E}_0^\vee \otimes \mathcal{G}) \to H^1(X, \mathcal{E}_1^\vee \otimes \mathcal{G})) $$
However, there is a surjection this may not be surjective
$$ \mathcal{E}_0^\vee \otimes \mathcal{G} \to \mathcal{E}_1^\vee \otimes \mathcal{G} \to 0 $$
and by vanishing of $H^2(X, -)$ the functor $H^1(X, -)$ is right exact so we see that,
$$ \mathrm{Ext}^2_{\mathcal{O}}(\mathcal{F}, \mathcal{G}) \cong \mathrm{coker}(H^1(X, \mathcal{E}_0^\vee \otimes \mathcal{G}) \to H^1(X, \mathcal{E}_1^\vee \otimes \mathcal{G})) = 0 $$
If $X$ is not separated it is not clear to me what to say about $\mathrm{Ext}^2$.
