8. 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 $\left[a_{n}\right]$ and $\left\{a_{n}\right\}$ represent the integer part and the fractional part of the real number $a_{n}$, respectively), then $a_{2019}=$
Given $a_{0}=1+\sqrt{3}-1, a_{1}=2+\frac{\sqrt{3}-1}{2}, a_{2}=4+\sqrt{3}-1, a_{3}=5+\frac{\sqrt{3}-1}{2}$, $a_{4}=7+\sqrt{3}-1, a_{5}=8+\frac{\sqrt{3}-1}{2}$, we can use mathematical induction to prove:
$$
a_{2 n}=3 n+1+\sqrt{3}-1, a_{2 n+1}=3 n+2+\frac{\sqrt{3}-1}{2} \text {. }
$$

Therefore, $a_{2019}=a_{2 \cdot 1009+1}=3 \cdot 1009+2+\frac{\sqrt{3}-1}{2}=\frac{6057+\sqrt{3}}{2}$.