A set is dense in
This is not dense. For example, the neighborhood with $r=1/3$ surrounding $(1/2,0)$ contains no points in this set (since $x\in\mathbb N$), so this point cannot be a limit point.
This is dense. It contains $\{(x,y):x,y\in \mathbb Q\}$ which is dense. The proof for its density is similar to the proof that $\mathbb Q$ is dense in $\mathbb R$.
This is not dense. The neighborhood surrounding the origin with $r=1$ contains no points in this set.
This is dense. Take $x,y \in \mathbb R$ such that $xy=0$. This is the complement of the set specified in the question with respect to $\mathbb R^2$. Then, $x=0$ or $y=0$.
Take a neighborhood around this set with radius $r$. Then, if $x=0$ and $y=0$, take the point $(r/2,r/2)$. This is a member of the neighborhood, so this point is a limit point.
If $x=0$ and $y\not =0$, then take the point $(r/2,y)$. This is a member of the neighborhood, so the point is a limit point.
Similarly, if $x\not=0$ and $y=0$, take the point $(x,r/2)$. The same argument as above shows that this is a limit point.