Usually when sets have such expressions it is very useful to check if working with continuous functions is possible, as it is in the case of $A$.
Note that if $f:\mathbb{R}^{2}\to \mathbb{R}$ is given by $f(x,y)=xy$, then $f$ is continuous and $A=f^{-1}(\mathbb{R}\setminus\{0\})$. Hence $A$ is open as a preimage of an open set under a continuous function. So $A$ equals its own interior. Moreover, for any point in the plane with either $x=0$ or $y=0$ the open ball $B((x,y),r)$ contains points $(a,b)$ such that $ab\neq 0$. Hence the boundary of $A$ consists of the $x$-axis and the $y$-axis. This leaves the exterioir of $A$ to be empty.
Note that $B=B(\bar{0},1)\cap \mathbb{Q}^{2}$, where $B(\bar{0},1)$ is the open $1$-radius ball around origin $\bar{0}$. So $B$ is basicly the rational coordinate points of the open unit ball. The interior of $B$ is empty since it contains no open balls: every open ball in $\mathbb{R}^{2}$ contains points with irrational coordinates and with rational coordinates. This being said, every open ball around a point in $C:=\{(x,y):x^{2}+y^{2}\leq 1\}$ has a non-empty intersection with $B$. This implies that $C$ is a subset of $B$'s boundary. Since the interior $B$ is empty then the boundary of $B$ equals its closure $\mathrm{cl}(B)$. Now since $C$ is closed and $B\subset C\subset \mathrm{cl}(B)$ it follows that $C$ is in fact the whole closure (since the closure is the smallest closed set containing $B$), and thus the boundary of $B$ is $C$ as well. Finally, the exterior of $B$ is what is left in $\mathbb{R}^{2}$, i.e. the set $\{(x,y):x^{2}+y^{2}>1\}$.