8. Given $99 \mid \overline{141 x 28 y 3}$, find $x, y$.

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8. Since $141 x 28 y 3 \equiv 0(\bmod 9)$, then
$$
141 x 28 y 3 \equiv 1+4+1+x+2+8+y+3 \equiv 0(\bmod 9) \text {, }
$$

we get $x+y+1 \equiv 0(\bmod 9)$, which means $x+y \equiv 8(\bmod 9)$,
thus we can set $x+y=9 k+8$, leading to $x+y=8$ or 17.
Since $141 x 28 y 3=0(\bmod 11)$, then
$$
141 x 28 y 3 \equiv 3+8+x+4-y-2-1-1 \equiv 0(\bmod 11) \text {, }
$$

we get $x-y \equiv 0(\bmod 11)$, thus we can set $x-y=11 l$,
leading to $x-y=0$, and since $0 \leqslant x, y \leqslant 9$, hence $x=y=4$ is the solution.