Describe the second cohomology group $H^2(Z_n times Z_n. k^*)$.

I would take the standard cyclic resolution of $G = Z/nZ$:
$$
dots stackrel{1-t}to Z[G] stackrel{sum t^i}to Z[G] stackrel{1-t}to Z[G] to Z to 0,
$$
where $t$ is the generator of $G$, and then take the tensor square of two such --- this would give a resolution
$$
dots to Z[G_1times G_2]^3 stackrel{d_2}to Z[G_1times G_2]^2 stackrel{d_1}to Z[G_1times G_2] to Z to 0,
$$
where $G_1 = G_2 = Z/nZ$ and the maps are given by
$$
d_1 = (1-t_1,1-t_2),
qquad
d_2 = left(begin{array}{ccc}
sum t_1^i & 1-t_2 & 0 cr
0 & 1-t_1 & sum t_2^i
end{array}right)
$$
($t_1$ and $t_2$ are the generators of $G_1$ and $G_2$ respectively).
I think you can use this for the calculations.