= Solution
After the $2n$ sides are identified in pairs, the <polygonal-schema Euler count> gives one face, $n$ edges and some number $V\geq1$ of vertices. The <Euler characteristic> of a closed orientable genus-$g$ surface is $2-2g$, so
$$
2-2g=V-n+1,
\qquad
n=V-1+2g\geq2g.
$$
For genus two, take a regular octagon in the <Poincare disc model> and identify each side with its opposite side, with the orientation reversed along the boundary. One cyclic labelling is
$$
a,b,c,d,a^{-1},b^{-1},c^{-1},d^{-1}.
$$
The quotient has $V=1$, $E=4$, and $F=1$, hence Euler characteristic $-2$ and genus two. All eight vertices become one point, so smoothness requires their angles to sum to $2\pi$. Each interior angle is consequently
$$
\boxed{\frac{2\pi}{8}=\frac\pi4}.
$$
Equivalently, the <hyperbolic polygon area> formula shows that the <regular hyperbolic octagon fundamental polygon> has area
$$
(8-2)\pi-8\left(\frac\pi4\right)=4\pi,
$$
as required by the <Gauss-Bonnet theorem> for a genus-two surface of curvature $-1$.
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