Ext groups of a point on a scheme Given a scheme $X$ over a field $k$ and a closed point $x$ with residue field $k(x)$ and inclusion $i:{x}\rightarrow X$ one can consider the following abelian groups
(1)$Ext^1_{\mathcal O_X}(i_*k(x),i_*k(x))$
and
(2)$Ext^1_{\mathcal O_{X,x}}(k(x),k(x))$.
The second one is just seen as Ext-group in the sense of modules over a ring.
Are they isomorphic?
I would define a map from (1) to (2) by just taking stalk in $x$, but I dont really see how one would come from (2) to (1).
Addition:
And is there a structure of a $k-$ vector space on (2) such that the iso is also one of $k-$spaces? (1) surely has a $k-$vector space structure as the scheme is defined over $k$.

The second Ext-group is an $\mathcal O_{X,x}$-module.  Since $\mathcal O_{X,x}$ is a $k$-algebra, it is in particular a $k$-vector space (and this is compatible with the $k$-v.s. structure on the first Ext-group and the map from the first to the second). 
Furthermore, the two Ext-groups are isomorphic (i.e. the map you defined is an isomorphism).  To see this, choose an affine open, say $U = $ Spec $A$, around $x$, and let $\mathfrak p$ be the prime ideal of $A$ corresponding to $x$.
Firstly, restriction to $U$ induces an isomorphism
$Ext^1_{\mathcal O_X}(k(x),k(x)) \cong Ext^1_{\mathcal O_U}(k(x),k(x))$
(since any extension of the two skyscrapers at $x$ is again supported on $x$,
and hence on $U$).  The second Ext group in this isomorphism can then be computed in terms of modules, and the isomorphism we want reduces to the isomorphism
$Ext^1_A(A/\mathfrak p, A/\mathfrak p) = Ext^1_{A_{\mathfrak p}}(A/\mathfrak p,A/\mathfrak p),$ which is straightforward.  (Any $A$-module which is an extension of $A_{\mathfrak p}$-modules is again an $A_{\mathfrak p}$-module.)
