I'm writing this as an "answer" because (a) there are a number of comments, and (b) I don't know if it would fit in a comment.
Let $F$ a finite field, and let $V$ a finite dimensional $F$-vector space, and view $V$ as an $F^\times$-module via multiplication. Then as pointed out in Andy Putman's answer, $H^i(F^\times,V) = 0$ for all $i \ge 0$ provided $|F| > 2$.
Well, it is clear enough under the assumption $|F|>2$ that $H^0(F^\times,V) = V^{F^\times} = 0$. For the higher cohomology vanishing, there is no need to use the description of "cohomology of cyclic groups" to obtain this vanishing; the point is just that $|F^\times|$ is invertible in $F$. Use the following generality:
Let $H$ be a subgroup of finite index $n$ in a group $G$. If $M$ is a $\mathbf{Z}G$-module, then $\operatorname{Cor} \circ \operatorname{Res}$ is multiplication by $n$ on $M$, where $\operatorname{Cor}:H^\bullet(H,M) \to H^\bullet(G,M)$ denotes the corestriction and $\operatorname{Res}:H^\bullet(G,M) \to H^\bullet(H,M)$ the restriction; see e.g. Serre's Local Fields VII.7, VIII.2.
Let now $k$ be a commutative ring (with 1), suppose that $H=1$ and that $n = [G:1]= |G|$ is invertible in $k$. If $M$ is a $kG$-module (i.e. a $k$-module with $k$-linear $G$ action), then all $H^i(G,M)$ are $k$-modules and $H^i(H,M) = H^i(1,M) = 0$ for $i>0$. For $i>0$, the preceding result shows these $k$-modules to be annihilated by the unit $n$ of $k$; thus $H^i(G,M) = 0$ for $i>0$.
To apply this result in the original setting, take $k=F$, $M=V$ and $G=F^\times$; we find that $H^i(F^\times,V) = 0$ for $i>0$.
