(5) Since $x_{n+1}=\frac{x_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} x_{n}}$, let $x_{n}=\tan a_{n}$, thus $x_{n+1}=\tan \left(a_{n}+\frac{\pi}{6}\right)$, $x_{n+6}=x_{n}$. It is easy to calculate that $x_{1}=1, x_{2}=2+\sqrt{3}, x_{3}=-2-\sqrt{3}, x_{4}=-1, x_{5}=$ $-2+\sqrt{3}, x_{6}=2-\sqrt{3}, x_{7}=1, \cdots$, so
$$
\sum_{n=1}^{2008} x_{n}=x_{1}+x_{2}+x_{3}+x_{4}=0 .
$$

Therefore, the answer is A.
5 Given the sequence $\left\{x_{n}\right\}, x_{1}=1$, E $x_{n+1}=\frac{\sqrt{3} x_{n}+1}{\sqrt{3}-x_{n}}$, then $\sum_{n=1}^{2008} x_{n}=(\quad)$.
(A) 0
(B) -1
(C) $2+\sqrt{3}$
(D) $-2+\sqrt{3}$