Recall that when a number is divisible by 3 then sum of its digits is also divisible by 3 .
So, we have
$4+6+9+5+3+x+y=3 k$ for some $k \in Z$
$\Rightarrow 27+(x+y)=3 k$
So, $(x+y)$ is also some multiple of 3 .
Since all the seven digit numbers is to have distinct therefore $x$ and $y$ must be chosen from $0,1,2,7,8$ such that $(x+y)$ is some multiple of 3 .
If $x=0$, then there is no possibility for $y$ If $x=1$, then $y=2,8 \therefore(1,2),(1,8)$ If $x=2$, then $y=1,7 \therefore(2,1),(2,7)$ If $x=7$, then $y=2,8 \therefore(7,2),(7,8)$ If $x=8$, then $y=1,7 \therefore(8,1),(8,7)$ $\therefore 8$ pairs are possible.