No. Take a f.g. non-finitely presented group $G$ with trivial $Out(G)$ and trivial center (such groups clearly exist). Suppose that $G$ is a normal subgroup of $H$. Then every $h\in H$ acts on $G$ by conjugation $x\to x^h$. Since $Aut(G)=Inn(G)$, there exists $g\in G$ such that $x^g=x^h$ for every $x\in G$. Hence $g^{-1}h$ centralizes $G$. Therefore $H=G Z_H(G)$ where $Z_H(G)$, the centralizer of $G$ in $H$, intersects $G$ trivially (since the center of $G$ is trivial) and is a normal subgroup. Hence $H$ is a direct product of $G$ and $Z=Z_H(G)$. Since $H$ is finitely generated, $Z$ is finitely generated (being a homomorphic image of $H$). Hence if $H$ is finitely presented, then $G$ is presented by the presentation of $H$ plus finitely many relations killing the generators of $Z$, so $G$ must be finitely presented, a contradiction.

No. Take a f.g. non-finitely presented group $G$ with trivial $Out(G)$ and trivial center (such groups clearly exist; in fact one can even assume that $Out(G)$ is locally finite, say $G$ is the Grigorchuk group of intermediate growth, by the result of Grigorchuk and Sidki). Suppose that $G$ is a normal subgroup of $H$. Then every $h\in H$ acts on $G$ by conjugation $x\to x^h$. Since $Aut(G)=Inn(G)$, there exists $g\in G$ such that $x^g=x^h$ for every $x\in G$. Hence $g^{-1}h$ centralizes $G$. Therefore $H=G Z_H(G)$ (if $Out(G)$ is locally finite, then instead of $G$ here you will get a finite extension of $G$, the image of $H$ in $Aut(G)$ under the natural homomorphism) where $Z_H(G)$, the centralizer of $G$ in $H$, intersects $G$ trivially (since the center of $G$ is trivial) and is a normal subgroup. Hence $H$ is a direct product of $G$ and $Z=Z_H(G)$. Since $H$ is finitely generated, $Z$ is finitely generated (being a homomorphic image of $H$). Hence if $H$ is finitely presented, then $G$ is presented by the presentation of $H$ plus finitely many relations killing the generators of $Z$, so $G$ must be finitely presented, a contradiction.

No. Take a f.g. non-finitely presented group $G$ with trivial $Out(G)$ and trivial center (such groups clearly exist). Suppose that $G$ is a normal subgroup of $H$. Then every $h\in H$ acts on $G$ by conjugation $\phi_g$. Since $Aut(G)=Inn(G)$, there exists $g\in G$ such that $x^g=x^h$ for every $x\in G$. Hence $g^{-1}h$ centralizes $G$. Therefore $H=G Z_H(G)$ where $Z_H(G)$ intersects $G$ trivially and is a normal subgroup. Hence $H$ is a direct product of $G$ and $Z=Z_H(G)$. Since $H$ is finitely generated, $Z$ is finitely generated. Hence if $H$ is finitely presented, then $G$ is presented by the presentation of $H$ plus finitely many relations killing the generators of $Z$, so $G$ must be finitely presented, a contradiction.

No. Take a f.g. non-finitely presented group $G$ with trivial $Out(G)$ and trivial center (such groups clearly exist). Suppose that $G$ is a normal subgroup of $H$. Then every $h\in H$ acts on $G$ by conjugation $x\to x^h$. Since $Aut(G)=Inn(G)$, there exists $g\in G$ such that $x^g=x^h$ for every $x\in G$. Hence $g^{-1}h$ centralizes $G$. Therefore $H=G Z_H(G)$ where $Z_H(G)$, the centralizer of $G$ in $H$, intersects $G$ trivially (since the center of $G$ is trivial) and is a normal subgroup. Hence $H$ is a direct product of $G$ and $Z=Z_H(G)$. Since $H$ is finitely generated, $Z$ is finitely generated (being a homomorphic image of $H$). Hence if $H$ is finitely presented, then $G$ is presented by the presentation of $H$ plus finitely many relations killing the generators of $Z$, so $G$ must be finitely presented, a contradiction.