Let $X$ be a topological space, $G\subseteq X$ be $G_{\delta}$ and $N\subseteq X$ be meagre in $X$ (and I don't know if this implies that $N$ is meagre in $G$).
Question 1. Is $G\setminus N$ dense in $G$?
Question 2. Can we find $N_{n}\subseteq X$ closed with non-empty interior, so that $N\subseteq\bigcup_{n\in\omega}N_{n}$, and so that $G\setminus\bigcup_{n\in\omega}N_{n}$ is dense in $G\setminus N$?
Question 1 is surely a resounding no, in general, even for nice spaces:
Take $X=\Bbb Q$, usual topology. $G=X$ is a $G_\delta$ and $N=X$ is meagre (union of countably many nowhere dense closed singletons). $G\setminus N=\emptyset$ is not dense in anything but $\emptyset$...
Or for another (Baire) space: $X=\Bbb R$, usual topology. $G=\Bbb Z$ which is a $G_\delta$ (as every closed set in a metrisable space) and also $N=G$ is nowhere dense so meagre. Again their difference is empty.
Note that these are also examples against 2 if you think about it.