8. Let $a, b$ be positive integers, and $a-b \sqrt{3}=(2-\sqrt{3})^{100}$, then the units digit of $a b$ is
Let $a_{n}-b_{n} \sqrt{3}=(2-\sqrt{3})^{n}$, then $a_{n+1}-b_{n+1} \sqrt{3}=(2-\sqrt{3})^{n+1}$
$$
=\left(a_{n}-b_{n} \sqrt{3}\right)(2-\sqrt{3})=2 a_{n}+3 b_{n}-\left(a_{n}+2 b_{n}\right) \sqrt{3} \Rightarrow\left\{\begin{array}{l}
a_{n+1}=2 a_{n}+3 b_{n}, \\
b_{n+1}=a_{n}+2 b_{n} .
\end{array}\right.
$$

Thus, $a_{1}=2, b_{1}=1 ; a_{2}=7, b_{2}=4 ; a_{3}=26, b_{3}=15 ; a_{4}=97, b_{4}=56 ; a_{5}=362$,
$$
b_{5}=209 ; a_{6}=1351, b_{6}=780 ; a_{7}=5042, b_{7}=2911 ; a_{8}=18817, b_{8}=10864 .
$$

Therefore, the unit digit of $a_{n} b_{n}$ is $2,8,0,2,8,0,2,8$, so the unit digit of $a b$ is 2.