8. If the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}=\left[(2+\sqrt{3})^{2^{n}}\right]\left(n \in \mathbf{Z}_{+}\right),
$$

then the last two digits of $a_{2020}$ are $\qquad$
8.53.

Let $b_{n}=(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}$. Then $b_{1}=4, b_{2}=14$, $b_{n+2}=4 b_{n+1}-b_{n}\left(n \in \mathbf{Z}_{+}\right)$.
Consider $\left\{b_{n}(\bmod 100)\right\}$, they are respectively:
$$
\begin{array}{l}
04,14,52,94,24,02,84,34,52,74,44, \\
02,64,54,52,54,64,02,44,74,52,34,84, \\
02,24,94,52,14,04,02,04,14, \cdots,
\end{array}
$$

It is a periodic sequence modulo 100 with a period of 30.
$$
\begin{array}{l}
\text { Also } 2^{2020} \equiv 2^{404}=2^{4} \times 2^{400} \equiv 2^{4} \times 2^{80} \\
\equiv 2^{4} \times 2^{16} \equiv 2^{5} \times 2^{15} \equiv 2 \times 2^{3} \\
\equiv 16(\bmod 30),
\end{array}
$$

Thus, the last two digits of $b_{2020}$ are 54.
Since $(2-\sqrt{3})^{n} \in(0,1)$, the last two digits of $a_{2020}$ are 53.