Is there a group $G$ and subgroup $H$, such that there exists $g\in G$ with $gHg^{-1} \subset H$ and $|H:gHg^{-1}|$ is infinite?

Let $\,G\,$ be the free group generated by $\,g,x_1,x_2,\dots\,$ modded by the equations $\,gx_ng^{-1} = x_{n+1}\,$ for all $n>0.$ Let $\,H\,$ be the subgroup generated by all the $\,x_n.\,$ The index of $\,gHg^{-1}\,$ in $\,H\,$ is infinite because $\,x_1\,$ has infinite order. Note that there are more concrete ways to represent the groups and specializations where all the $\,x_n\,$ commute with each other but not with $\,g.\,$

Yes.

For example, take $G$ to be the HNN-extension* $$\langle a, b, t\mid t^{-1}at=[a, b], t^{-1}bt=b\rangle$$ and $H=\langle a, b\rangle$. By the theory of HNN-extensions, $H$ embeds into $G$ in the natural way. Clearly $t^{-1}Ht\leq H$. To see that $|H:t^{-1}Ht|$, note that the presentation $\langle a, b\mid [a, b], b\rangle$ defines an infinite group. Hence the smallest normal subgroup of $H$ containing $[a, b]$ and $b$ has infinite index in $H$; hence the smallest subgroup of $H$ containing $[a, b]$ and $b$, which is $\langle [a, b], b\rangle$, has infinite index in $H$. As $t^{-1}Ht=\langle [a, b], b\rangle$, the result follows.

*For background on HNN-extensions see Wikipedia or the book Combinatorial group theory by Lyndon and Schupp.

$\newcommand{\bZ}{\mathbf Z}\newcommand{\bN}{\mathbf N}$ Let $\overline{H}$ be the group of functions $\bZ\to \bZ$ (with pointwise addition), while $H$ is the group of functions which vanish on negative integers.

Now, consider the semidirect product $G:=\overline{H}\rtimes \bZ$, where $\bZ$ acts on $\overline{H}$ by shifting to the right, so $(1\cdot f)(k)=f(k+1)$. Then if you take $g=(0,1)\in G$, then $gHg^{-1}\unlhd H$ (it is just the set of functions $\bZ\to \bZ$ which vanish on non-positive integers) and $H/gHg^{-1}\cong \bZ$.

If you replace functions $\bZ\to \bZ$ with functions into any other group $K$, then you will have $H/gHg^{-1}\cong K$.

Note that this construction is the (unrestricted) wreath product of $\bZ$ with itself.