Twisted cohomology of torus I think I could write down a projective resolution, tensor with the twisted coefficients and find the first cohomology of the standard torus.
BUT, I was wondering if there is an easier way to understand the first (co)homology groups. (Just checking, by Poincaré duality they should be isomorphic, right?)
Is there something like a Künneth formula for twisted coefficients?
 A: Yes, under suitable circumstances there is a Künneth formula for local coefficients. This can be found in Theorems 1.6, 1.7 of the following paper:
R. Greenblatt: Homology with local coefficients and characteristic classes. Homology, Homotopy and Applications, 8(2), 2006, 91-103. 
But I agree with Fernando that in case of a torus it is easier to construct a projective resolution. In particular, this way you can handle any twist, while the theorems cited above only adress "product actions". 
A: Let me offer you a projective resolution of $\mathbb{Z}$ as a module over $\pi_1(T)=\mathbb{Z}^2$, obtained from the cellular chain complex of $\mathbb{R}^2$, the universal cover of $T$. The cell structure on $\mathbb{R}^2$ is induced by the usual cell structure on $T$ with one vertex, two edges and one $2$-cell.
The group ring is the ring of Laurent polynomials in two variables $R=\mathbb{Z}[x^{\pm 1},y^{\pm 1}]$. The resolution is
$$0\rightarrow R\stackrel{({y-1},{1-x})^t}\longrightarrow R^2\stackrel{(x-1,y-1)}\longrightarrow R\rightarrow \mathbb{Z}\rightarrow 0$$
