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 $\mathrm{Ext}^2$.
Best Answer
If $X$ is smooth then $$ \mathrm{Ext}^i(F,F) \cong \mathrm{H}^i(X, \mathrm{R}\mathcal{H}\mathit{om}(F,F)) \cong \mathrm{H}^i(X, \mathrm{R}\mathcal{H}\mathit{om}(F,\mathcal{O}_X) \otimes^{\mathrm{L}} F), $$ so it is enough to construct a local derived trace morphism $$ \mathrm{R}\mathcal{H}\mathit{om}(F,\mathcal{O}_X) \otimes^{\mathrm{L}} F \to \mathcal{O}_X\tag{*} $$ and then apply cohomology. For this just note that the functor $$ G \mapsto \mathrm{R}\mathcal{H}\mathit{om}(F,G) $$ is right adjoint to the functor $$ G \mapsto G \otimes^{\mathrm{L}} F, $$ and $(*)$ is just the counit of adjunction applied to $\mathcal{O}_X$.