For a finite field $\mathbb{F}$ of char $p$ $\geq 2$, let $f$ be a normalised eigenform of weight $k\geq 2$ that is ordinary at $p$ and $\overline{\rho}_f$ be the mod $p$ Gal representation attached to it. Let $G_p$ and $I_p$ be the decomposition and inertia subgroups. We know that the restriction of $\overline{\rho}_f$ to $G_p$ is uppertriangular with characters $\delta$ and $\epsilon$ (unramified) on the diagonal. Under the assumption that $\delta$ and $\epsilon$ are distinct mod $p$ ($p$distinguishedness) why is it true that $H^1(G_p/I_p,\mathbb{F}(\delta\epsilon^{1})) = 0$? I know that $G_p/I_p \simeq \hat{\mathbb{Z}}$ has cohomological dimension $1$ but this only shows that the cohomology groups vanish from $H^2$ onwards . Is it to do with the fact that $G_p/I_p$ is topologically generated? Could someone please explain this to me?
A variant of David Loeffler's answer... If more generally $V$ is a $G_{\mathbb Q_{p}}$representation with coefficients in a field, then the dimension of $H^{1}(G_{p}/I_{p},V)$ (EDIT : or rather $H^{1}(G_{p}/I_{p},V^{I_{p}})$) is equal to the dimension of $H^{0}(G_{p},V)$ because the former is the cokernel of the map $Fr_{p}1$ acting on $V^{I_{p}}$ and the latter is the kernel of this map. In the situation you described, $\mathbb F(\delta\epsilon^{1})$ has no nontrivial invariants under $G_{p}$ because $\delta≠\epsilon$ (and this is why you need your form to be $p$distinguished) so its $H^{1}$ is trivial as well. 


Cohomology of $\widehat{\mathbb{Z}}$modules is something you can just write down. If $V$ is a (finite) module with a (continuous) action of $\widehat{\mathbb{Z}}$, so the generator acts by some endomorphism $\varphi$, then the $H^0$ is the $\varphi$invariants (obviously), and the $H^1$ is the coinvariants $V / (1  \varphi)V$. If $V$ is $\mathbb{F}_p$, with $\varphi$ acting by an element $u \ne 1 \in \mathbb{F}_p^\times$, then $1\varphi$ is a nonzero linear map from $\mathbb{F}_p$ to itself; so it's surjective, and the $H^1$ is zero. 

