8.34 The real number sequence $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfies the following equation: $a_{0}=a$, where $a$ is a real number,
$$a_{n}=\frac{a_{n-1} \sqrt{3}+1}{\sqrt{3}-a_{n-1}}, n \in N .$$

Find $a_{1994}$.
［Solution］ After calculation, we can obtain
$$\begin{aligned}
a_{1} & =\frac{a \sqrt{3}+1}{\sqrt{3}-a} \\
a_{2} & =\frac{a_{1} \sqrt{3}+1}{\sqrt{3}-a_{1}}=\frac{(a \sqrt{3}+1) \sqrt{3}+(\sqrt{3}-a)}{\sqrt{3}(\sqrt{3}-a)-(a \sqrt{3}+1)} \\
& =\frac{a+\sqrt{3}}{1-a \sqrt{3}} \\
a_{3} & =\frac{a_{2} \sqrt{3}+1}{\sqrt{3}-a_{2}}=\frac{(a+\sqrt{3}) \sqrt{3}+(1-a \sqrt{3})}{\sqrt{3}(1-a \sqrt{3})-(a+\sqrt{3})} \\
& =-\frac{1}{a}
\end{aligned}$$

From this, we get $\square$
$$a_{6}=-\frac{1}{-\frac{1}{a}}=a$$

It is evident that the sequence $a_{0}, a_{1}, a_{2}, \cdots$ is a periodic sequence with a period of 6.
Since $1994=6 \times 332+2$, we have
$$a_{1994}=a_{2}=\frac{a+\sqrt{3}}{1-a \sqrt{3}} .$$

Where $k, u$ are both positive integers.
Substituting $k=u u_{5}$, we get
$$\begin{array}{l}
u^{2} \cdot u_{5}=2000 \\
u^{2} \cdot u_{5}=2^{4} \cdot 5^{3}
\end{array}$$
All possible values of $u^{2}$ are $1,2^{2}, 2^{4}, 5^{2}, 2^{2} \cdot 5^{2}, 2^{4} 5^{2}$, thus all possible values of $u$ are 1, $2,2^{2}, 5,2 \cdot 5,2^{2} \cdot 5$; all possible values of $k$ are $2000,1000,500,400,200,100$.