8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. The sequence $\left\{a_{n}\right\}$ satisfies:
$$
x_{1}=1, x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \text {. }
$$

Then the units digit of $x_{2021}$ is
8.9.

Given $x_{2}=7$.
Notice, for any $x_{n} \in \mathbf{Z}_{+}$,
$$
\begin{array}{l}
4 x_{n}+\sqrt{11} x_{n}-1<x_{n+1}<4 x_{n}+\sqrt{11} x_{n} \\
\Rightarrow 5 x_{n}-(4-\sqrt{11})<(4-\sqrt{11}) x_{n+1}<5 x_{n} \\
\Rightarrow\left\{\begin{array}{l}
4 x_{n+1}-5 x_{n}<\sqrt{11} x_{n+1}, \\
\sqrt{11} x_{n+1}+(4-\sqrt{11})<4 x_{n+1}-5 x_{n}
\end{array}\right. \\
\Rightarrow 4 x_{n+1}-5 x_{n}=\left[\sqrt{11} x_{n+1}\right] \\
\Rightarrow x_{n+2}=4 x_{n+1}+\left[\sqrt{11} x_{n+1}\right]=8 x_{n+1}-5 x_{n} \text {. } \\
\end{array}
$$

Clearly, $x_{n+2} \equiv x_{n}(\bmod 2)$,
$$
x_{n+2} \equiv 3 x_{n+1}(\bmod 5) \text {. }
$$

Given $x_{1}=1, x_{2}=7$, it is easy to prove by mathematical induction that for any $n \in \mathbf{Z}_{+}, x_{n} \equiv 1(\bmod 2)$.
Combining $x_{1}=1, x_{2}=7$, we get when $n \geqslant 2$,
$$
x_{n} \equiv 2 \times 3^{n-2}(\bmod 5) \text {. }
$$

Thus, $x_{2021} \equiv 2 \times 3^{2019} \equiv 2 \times 3^{3} \equiv 4(\bmod 5)$. Therefore, the required unit digit is 9.