Let $H$ be a normal subgroup of order $6$. If $f :G\longrightarrow G_1$ be an epimorphism of groups such that $H\subset \mathrm{Ker} (f)$, then show that $G_1$ is also a homomorphic image of $G/H$.
In this problem, from the first isomorphism theorem, we can say that $G/\mathrm{Ker}(f)\cong G_1$. Hence, there exists an isomorphism $f:G/\mathrm{Ker}(f)\longrightarrow G_1$, such that $f$ is bijective. Given, $H\subset \mathrm{Ker}( f)$, if we consider the same function $f:G/H\longrightarrow G_1$, then we can probably think this situation where, the original domain of $f$ have been shrinked and this change keeps the property of $f$ invariant except the fact $f$ does not remain onto. But $f$ is still a homomorphism, due to which the fact that $G_1$ is also a homomorphic image of $G/H$ follows trivially.
Can this considered a valid solution? If not, then I need an explicit reason for the same. I know there are posts associated with this particular topic in this thread, but I want to know whether this particular solution is valid or, not?
Your solution is flawed, at the least, in that once again the expression "homomorphic image" means a surjective homomorphism. If you were just looking for any homomorphism, there's always the trivial one.
This follows from the universal property of quotient groups. See here.
There's a unique homomorphism $\bar f:G/H\twoheadrightarrow G_1, $ where $f=\bar f\circ \pi$, and $\pi$ is the canonical projection onto $G/H.$
I don't see any need that $H$ have order $6$.