= Solution
\b[False.] Use the two-generated <group> $G_S$ constructed in part (iii) for a non-recursively-enumerable $S$. Sending the two free generators to $b_0,t$ gives a surjective <group homomorphism>
$$
\boxed{F_2\twoheadrightarrow G_S.}
$$
The source is a <finitely presented group>, with presentation $\langle a,b\mid\ \rangle$. The target is not finitely presented: if it were, its null words could be enumerated from a finite set of relators, and the tested commutators would enumerate $S$, as proved in part (iii). Thus even a quotient of a rank-two <free group> need not admit a finite presentation. A quotient by a finitely normally generated kernel would be finitely presented, but the assertion places no such restriction on its kernel.
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