8. Given
$$
a_{n}=\mathrm{C}_{200}^{n}(\sqrt[3]{6})^{200-n}\left(\frac{1}{\sqrt{2}}\right)^{n}(n=1,2, \cdots, 95) \text {. }
$$

The number of integer terms in the sequence $\left\{a_{n}\right\}$ is $\qquad$
8. 15 .

Notice that $a_{n}=\mathrm{C}_{200}^{n} \times 3^{\frac{200-n}{3}} \times 2^{\frac{200-5 n}{6}}$.
To make $a_{n}(1 \leqslant n \leqslant 95)$ an integer, it must be that $\frac{200-n}{3}$ and $\frac{400-5 n}{6}$ are both integers, i.e., $61(n+4)$.

When $n=6 k+2(k=0,1, \cdots, 13)$, $\frac{200-n}{3}, \frac{400-5 n}{6}$ are both non-negative integers. Therefore, $a_{n}$ is an integer, with a total of 14.

When $n=86$, $a_{86}=\mathrm{C}_{200}^{86} \times 3^{38} \times 2^{-5}$, in $\mathrm{C}_{200}^{86}=\frac{200!}{86!\times 114!}$, the number of factor 2 in $200!$ is $\left[\frac{200}{2}\right]+\left[\frac{200}{2^{2}}\right]+\cdots+\left[\frac{200}{2^{7}}\right]=197$.
Similarly, the number of factor 2 in 86! is 82, and the number of factor 2 in 114! is 110.
Thus, the number of factor 2 in $\mathrm{C}_{200}^{86}$ is
$197-82-110=5$.
Therefore, $a_{86}$ is an integer.
When $n=92$, $a_{92}=\mathrm{C}_{200}^{92} \times 3^{36} \times 2^{-10}$.
Similarly, the number of factor 2 in $\mathrm{C}_{200}^{92}$ is less than 10.
Therefore, $a_{92}$ is not an integer.
Thus, the number of integer terms is $14+1=15$.