$\newcommand{\Zhat}{\widehat{\mathbb{Z}}}$

Here are some not-to-difficult conclusions, though I would welcome any additional input.

There is a natural map $$p : F\rightarrow F^\Delta\times\Zhat^2$$ The kernel of $F\rightarrow\Zhat^2$ is just $[F,F]$, and since all finite quotients of $F^\Delta$ are perfect, $[F,F]$ surjects onto $F^\Delta$, and thus the map $p$ above is surjective. Since the all finite simple quotients of $F$ must factor through $F^\Delta\times\Zhat^2$, the kernel of $p$ is contained in the intersection of all maximal normal subgroups of $F$, otherwise known as the "Jacobson radical" of $F$ (also known as the Baer radical, or small radical). This is similar to the Frattini subgroup $\Phi(F)$, but in this case it cannot equal to $\Phi(F)$, since by Corollary 8.7.5 in Ribes/Zalesskii, $\Phi(F) = 1$ (thanks to Benjamin Steinberg for the reference), and by Proposition 8.7.7, $F$ is not a direct product.

Nonetheless, we get maps $\newcommand{\Aut}{\text{Aut}}$ $$\Aut(F)\twoheadrightarrow\Aut(F^\Delta\times\Zhat^2)\cong \Aut(F^\Delta)\times\Aut(\Zhat^2)$$ where the isomorphism is due to the fact that the projections of $F^\Delta\times\Zhat^2$ onto each direct factor is a characteristic quotient.

Thus, if $G$ is a 2-generated pro-$\Delta$ group, then any surjection $F\rightarrow G$ must factor through $F^\Delta$. This implies a positive answer to the first question if ENA = $\Delta$ and $G$ is pro-$\Delta$. Actually, by similar arguments, as long as $F/K_G$ is perfect, the answer to the first question is also positive.

This also gives a positive answer to the second question.

For the third question, the above shows that the map is definitely surjective, though not injective - for example, inner automorphisms by elements in the Jacobson radical of $F$ lie in the kernel.

$\newcommand{\Zhat}{\widehat{\mathbb{Z}}}$

Here are some partial answers, though I would welcome any additional input.

There is a natural map $$p : F\rightarrow F^\Delta\times\Zhat^2$$ The kernel of $F\rightarrow\Zhat^2$ is just $[F,F]$, and since all finite quotients of $F^\Delta$ are perfect, $[F,F]$ surjects onto $F^\Delta$, and thus the map $p$ above is surjective. Since the all finite simple quotients of $F$ must factor through $F^\Delta\times\Zhat^2$, the kernel of $p$ is contained in the intersection of all maximal normal subgroups of $F$, otherwise known as the "Jacobson radical" of $F$ (also known as the Baer radical, or small radical). This is similar to the Frattini subgroup $\Phi(F)$, but in this case it cannot equal to $\Phi(F)$, since by Corollary 8.7.5 in Ribes/Zalesskii, $\Phi(F) = 1$ (thanks to Benjamin Steinberg for the reference), and by Proposition 8.7.7, $F$ is not a direct product.

Nonetheless, we get maps $\newcommand{\Aut}{\text{Aut}}$ $$\Aut(F)\twoheadrightarrow\Aut(F^\Delta\times\Zhat^2)\cong \Aut(F^\Delta)\times\Aut(\Zhat^2)$$ where the isomorphism is due to the fact that the projections of $F^\Delta\times\Zhat^2$ onto each direct factor is a characteristic quotient.

Thus, if $G$ is a 2-generated pro-$\Delta$ group, then any surjection $F\rightarrow G$ must factor through $F^\Delta$. This implies a positive answer to the first question if ENA = $\Delta$ and $G$ is pro-$\Delta$. Actually, by similar arguments, as long as $F/K_G$ is perfect, the answer to the first question is also positive.

This also gives a positive answer to the second question.

For the third question, the above shows that the map is definitely surjective, though not injective - for example, inner automorphisms by elements in the Jacobson radical of $F$ lie in the kernel.