Solvable irreducible subgroups of the $\mathbf{GL}_n$ of $\mathbf{F}_p$ ($p$ prime)

Here's a slightly different answer, less group-theoretic and more representation-theoretic than Geoff's.

Rephrasing your question in terms of $\mathbb{F}_pG$-modules, you are asking about a faithful irreducible $\mathbb{F}_pG$-module $E$ for a finite solvable group $G$.

Let $N$ be a minimal normal subgroup of $G$. Since $G$ is solvable, $N$ is an elementary abelian $q$-group for some prime $q$. By Clifford's Theorem, the restriction of $E$ to $N$ is semisimple.

If $q=p$, then the restriction of $E$ to $N$ is trivial, since the only simple $\mathbb{F}N$-module for a $p$-group $N$ is the trivial module. But this contradicts the faithfulness of $E$.

If $q\neq p$ then $e_N=\frac{1}{\vert N\vert}\sum_{n\in N}n$ is a central idempotent of $\mathbb{F}_pG$. Let $f_N=1-e_N$.

$E=Ee_N\oplus Ef_N$, and since $E$ is irreducible, either $E=Ee_N$ or $E=Ef_n$.

If $E=Ee_N$ then, since the trivial $\mathbb{F}_pN$-module is the only irreducible $\mathbb{F}_pN$-module which is not annihilated by $e_N$, $N$ acts trivially on $E$, again contradicting the faithfulness of $E$.

So $E=Ef_N$, which means that $E$ and the trivial module $\mathbb{F}_p$ are in different blocks of $\mathbb{F}_pG$, and so $H^i(G,E)=\operatorname{Ext}^i_{\mathbb{F}_pG}(\mathbb{F},E)=0$ not only for $i=1,2$ but for all $i$.

It is true that if $M$ is any (solvable) group with $E \lhd M$ and $M/E \cong G$, (with the action of $G$ on $E$ specified by the given irreducible module action), then $E$ is complemented in $M$, and all complements to $E$ are conjugate. In fact, this does not require solvability of $G,$ only $p$-solvability. All this is well-known, but I outline the proof: Note that $O_{p}(G) = 1$ since $G$ acts irreducibly on $E$, so that $O_{p}(M) = E.$ Let $K = O_{p,p^{\prime}}(M)$. Then $K$ has a unique conjugacy classes of Hall $p^{\prime}$-subgroups, say one of these is $L$, so by a Frattini-like argument we have $M = KN_{M}(L)$.

Hence $M = ELN_{M}(L) = EN_{M}(L).$ Now $E$ is a minimal normal subgroup of $M$ since $G$ acts irreducibly on $E.$ However $E \cap N_{M}(L) \lhd M$. We can't have $E \leq N_{M}(L)$, otherwise $[E,L] = 1,$ whereas $G$ acts faithfully on $E$. Hence $E \cap N_{M}(L) = 1$ and $G \cong N_{M}(L)$. This argument actually shows that every complement to $E$ in $M$ has the form $N_{M}(L)$ for some Hall $p^{\prime}$-subgroup $L$, of $O_{p,p^{\prime}}(M)$ and $L$ is unique up to $M$-conjugacy, so all complements to $E$ are $M$-conjugate.