Some suggested directions- not a complete answer :We may,and do, suppose that $H_{1} \cap H_{2}$ is corefree. Let $[G:H_{1}] = p^{a}$ for some prime $p$ and integer $a.$ Then $O_{p^{\prime}}(G) = 1$ as $O_{p^{\prime}}(G) \leq H_{1} \cap H_{2}.$ 
  We note the general fact that if $M$ is a subgroup of the finite group $G,$ and $[G:M]$ is a power of $p,$ then $O_{p}(M) \leq O_{p}(G).$ It suffices to prove that $O_{P}(M) \leq P$ whenever $P \in {\rm Syl}_{p}(G).$ But $|PM| = \frac{|P||M|}{|P \cap M|}$, and this is divisible both by  $|G|_{p^{\prime}}|$ and by $|P|,$ so that $PM = G$ and $P \cap M \in {\rm Syl}_{p}(M)$. Hence $O_{p}(M) \leq P \cap M,$ as claimed.
   Hence we have $O_{p}(H_{i}) \leq O_{p}(G)$ for each $i.$ Note that $|O_{p}(H_{1})| = |O_{p}(H_{2})|.$ 
   Now $O_{p}(G) \not \leq H_{1},$ for otherwise $O_{p}(H_{2}) \leq O_{p}(G) \leq O_{p}(H_{1}) \leq O_{p}(G) $ and $O_{p}(H_{1}) = O_{p}(H_{2}) = O_{p}(G)$ contrary to the assumption that $H_{1} \cap H_{2}$ is corefree.
   Hence $G = O_{p}(G)H_{1}$ and $$|G| = \frac{|O_{p}(G)||H_{1}|}{O_{p}(H_{1})| |G| = \frac{|O_{p}(G)||H_{2}|}{O_{p}(H_{2})|,$ so $G= O_{p}(G)H_{2}.$ (to be continued)