= Solution
We prove <Hall subgroup existence in soluble groups> by induction on $|G|$. The trivial group is immediate. Choose a nontrivial minimal <normal subgroup> $N$, <elementary abelian> of order $p^a$ by part (c). Induction gives a Hall $\pi$-subgroup of $G/N$; let $K$ be its full preimage.
If $p\in\pi$, then $K$ itself is the required subgroup. If $p\notin\pi$ and $K<G$, apply induction inside the soluble subgroup $K$ to obtain a Hall $\pi$-subgroup $H$ of $K$. Since $[G:K]$ and $[K:H]$ are both $\pi'$-numbers, $H$ is Hall in $G$ too.
It remains to treat $p\notin\pi$ and $K=G$. Then $G/N$ is a $\pi$-group. If $G=N$, the subgroup one works. Otherwise choose a minimal <normal subgroup> $M/N$ of $G/N$, an <elementary abelian> $q$-group with $q\in\pi$. Let $Q$ be a Sylow $q$-subgroup of $M$. As $|M|=|N|q^b$ and $p\ne q$, we have $M=NQ$. The permitted <Frattini argument> gives
$$
G=N_G(Q)M=N_G(Q)N.
$$
The last equality uses $Q\leq N_G(Q)$ and the normality of $N$.
If $T=N_G(Q)<G$, then $[G:T]=[N:N\cap T]$ is a power of $p$, hence a $\pi'$-number. Apply induction to $T$ and multiply indices as before. If $T=G$, then $Q$ is a nontrivial normal $\pi$-subgroup. Induction in $G/Q$ gives a Hall $\pi$-subgroup whose full preimage in $G$ is Hall, since its additional factor $|Q|$ is a $\pi$-number. These cases exhaust the possibilities, proving \b[existence for every prime set]. No unproved complement theorem or conjugacy theorem for Hall subgroups was inserted into the proof.
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