Let $A$ be a set in a topological space $X$. We know that $A$ is dense in its closure $\bar{A}$. This implies that $\bar{A} \setminus A$ is nowhere dense in $\bar{A}$ (using a characterisation of nowhere dense sets in Kechris' Classical Descriptive Set Theory, section 8.A, namely that a set is nowhere dense if and only if its complement is dense). Since every nowhere dense set is also meagre, we know that $\bar{A} \setminus A$ is meagre and therefore $\bar{A} \setminus (\bar{A} \setminus A) = A$ is comeagre in $\bar{A}$.
Does this argument work?
Edit: Just noticed that I misread Kechris' characterisation. It is that a set $A$ is nowhere dense if and only if $X \setminus \bar{A}$ is dense (i.e. not when its complement is dense). Then obviously, my argument doesn't work.
No, for instance take $\mathbb{Q}$. It is meager in its closure $\mathbb{R}$.
So your third sentence is already false, although it holds for open $A$.