Is the cohomology of a stalk the same as the stalk of the cohomology sheaf? Let $\mathcal F^\bullet$ be a cochain complex of sheaves, then
$$H^i(\mathcal F^\bullet)(U)=\frac{\ker (d : \mathcal F^i(U)\to \mathcal F^{i+1}(U))}{\mathrm{im} (d: \mathcal F^{i-1}(U)\to \mathcal F^{i}(U))}$$
and
$$H^{i}(\mathcal F^\bullet_x) = \frac{\ker (d : \mathcal F^i_x\to  \mathcal F^{i+1}_x)}{\mathrm{im} (d: \mathcal F^{i-1}_x\to \mathcal F^{i}_x)}$$
are defined.

Is $H^{i}(\mathcal F^\bullet)_x$ canonically isomorphic to $H^i(\mathcal F^\bullet_x)$?

As a counter example I was thinking about a sheaf complex on $[0,1]$ that is the constant sheaf in degree $0$ and a sub-sheaf of $C[0,1]$ in degree $1$ that contains only the restrictions and linear combinations of a class of functions $f_n$ with support in $[1/n,1]$. The image of $f_n$ under $d$ should be, on connected components, the constant generator. No other degrees are non-zero.
The stalk $\mathcal F^1_0$ is I think zero, thus the homology of the stalk in degree $0$ should be the coefficients $\Bbb R$. However the homology sheaf is $0$ in degree $0$, and so the stalk of the homology sheaf should also be $0$.
I think my example can work, but I'm not so sure about a few details.
 A: Yes: the functor sending a sheaf to its stalk at $x$ is an exact functor, so it preserves kernels, cokernels, homology, etc.  Since taking stalks is clearly additive, it suffices to show it preserves kernels and cokernels.  To prove this, first note that kernels and cokernels of presheaves are computed pointwise (i.e. the value of a (co)kernel presheaf on an open set is just the (co)kernel of the map of sections over that open set).  The stalk of a presheaf is a filtered colimit of its values on open sets, and filtered colimits preserve kernels and cokernels.  So taking stalks of presheaves preserves kernels and cokernels.  It follows immediately that taking stalks of sheaves preserves kernels, since kernels of sheaves are the same thing as kernels of presheaves.  To compute a cokernel of sheaves, you first take the cokernel of presheaves and then sheafify.  Since sheafification does not change stalks, the stalk of the cokernel sheaf is the same as the stalk of the cokernel presheaf which is the same as the cokernel of the stalks.  Thus taking stalks of sheaves preserves cokernels.
I don't entirely follow your proposed counterexample (in particular, I don't understand what the differential is supposed to be), but I don't see how it could possibly have $\mathcal{F}^1_0=0$ as you claim.  Doesn't each $f_n$ have a germ at $0$?  The image of $f_n$ in $\mathcal{F}^1_0$ will be nonzero unless $f_n$ is identically zero in a neighborhood of $0$.
