See http://mathoverflow.net/questions/75472/kuenneth-formula-for-group-cohomology-with-nontrivial-action-on-the-coefficient

Let $C=C_2$ be the cyclic group of order two, $\def\ZZ{\mathbb Z}\ZZ$ the trivial module over $C$ and $S$ the $C$-module which is $\ZZ$ as an abelian group with with the non-trivial action of $C$. You want to compute $H^\bullet(C\times C,M)$ with $M=\ZZ\otimes S$, and the Künneth formula tells us that there is a short exact sequence $$0\to (H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^p\to H^p(C\times C,M)\to (Tor^\ZZ_1(H^\bullet(C,\ZZ),H^\bullet(C,S))^{p+1}\to0$$

You can easily compute $H^\bullet(C,-)$ with coefficients in every $C$-module and the Tor is also easy to compute. If I am not making too many mistakes, then \begin{gather}
(H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^{p}=(Tor_1(H^\bullet(C,\ZZ),H^\bullet(C,S))^{p}=0
\end{gather}
for all even $p\geq0$, and
\begin{gather}
(H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^{p}=(\ZZ/2\ZZ)^r,\\
(Tor_1(H^\bullet(C,\ZZ),H^\bullet(C,S))^{p}=(\ZZ/2\ZZ)^{r-1}
\end{gather}
for all odd $p=2r-1\geq0$. If follows that in the short exact sequenc above exactly one of the first and third terms are not zero for a given $p$, so we get an isomorphism in all cases.