8. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{0}=\sqrt{3}, a_{n+1}=\left[a_{n}\right]+\frac{1}{\left\{a_{n}\right\}},
$$

where, $[x]$ denotes the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Then $a_{2019}=$ $\qquad$
8. $3029+\frac{\sqrt{3}-1}{2}$.

Notice that, $a_{0}=1+(\sqrt{3}-1)$,
$$
\begin{array}{l}
a_{1}=1+\frac{1}{\sqrt{3}-1}=2+\frac{\sqrt{3}-1}{2}, \\
a_{2}=2+\frac{2}{\sqrt{3}-1}=3+\sqrt{3}=4+(\sqrt{3}-1), \\
a_{3}=4+\frac{1}{\sqrt{3}-1}=5+\frac{\sqrt{3}-1}{2} .
\end{array}
$$

By induction we get
$$
\begin{array}{c}
a_{2 k}=3 k+1+(\sqrt{3}-1), \\
a_{2 k+1}=3 k+2+\frac{\sqrt{3}-1}{2} .
\end{array}
$$

Therefore, $a_{2019}=3029+\frac{\sqrt{3}-1}{2}$.