Define a function $f_{2}, Q \rightarrow C$ that satisfies the following conditions:
(1) For any 1988 rational numbers $x_{1}, x_{2}, \cdots, x_{1988}$, we have
$$
\begin{array}{l}
f\left(x_{1}+x_{2}+\cdots+x_{1988}\right) \\
=f\left(x_{1}\right) \cdot f\left(x_{2}\right) \cdot \cdots \cdot f\left(x_{1988}\right) ;
\end{array}
$$
(2) For any rational number $x$, we have
$$
f(1988) f(x)=f(1988) \bar{f}(x) \text {. }
$$

Find all such functions.
Obviously, $f \equiv 0$ is a solution. Suppose $f \neq 0$, then there exists a rational number $x_{0}$, such that $f\left(x_{0}\right) \neq 0$. In (1), let $x_{1}=x_{0}, x_{2}=x_{3}=\cdots=x_{1883}=0$, we get
$$
f\left(x_{0}\right)=f\left(x_{0}\right) \cdot(f(0))^{1987} \text {. }
$$

Since $f\left(x_{0}\right) \neq 0$, we have
$$
(f(0))^{1987}=1 .
$$

Let $\omega=f(0)$, then $\omega^{1987}=1$.
Let $g(x)=f(x) / \omega$, then $g(0)=1$, condition (1) becomes
$$
\begin{array}{l}
g\left(x_{1}+x_{2}+\cdots+x_{1988}\right) \\
=g\left(x_{1}\right) g\left(x_{2}\right) \cdots g\left(x_{1988}\right) ;
\end{array}
$$

Condition (2) becomes
$$
\overline{g(1988)} \cdot g(x)=g(1988) \cdot \bar{g}(x) .
$$

In (3), let $x_{3}=x_{4}=\cdots=x_{1988}=0$, using $g(0)=1$, we get
$$
g\left(x_{1}+x_{2}\right)=g\left(x_{1}\right) \cdot g\left(x_{2}\right) .
$$

If there exists $x^{\prime}$ such that $g\left(x^{\prime}\right)=0$, then for any $x \in Q$, by (5) we have
$$
g(x)=g\left(x^{\prime}\right) \cdot g\left(x-x^{\prime}\right)=0,
$$

especially, $g(0)=0$. This contradicts the previous assumption. Therefore, for any rational number $x, g(x) \neq 0$.
$$
\begin{array}{l}
\text { In (4), let } x=0, \text { by } g(0)=1 \text { we get } \\
g(1988)=g(1988) .
\end{array}
$$

Thus, (4) can be further simplified to
$$
g(x)=u(x),
$$

i.e., $g(x)$ is a real function.
$$
\begin{array}{l}
\text { In (5), let } x_{1}=x_{2}=x, \text { then } \\
g(x)=\left[g\binom{x}{2}\right]^{2}>0 . \\
\end{array}
$$
For any rational number $\frac{m}{n}$, we have
$$
\begin{array}{l}
=g\binom{1}{n} \cdot g\binom{1}{n} \cdot \cdots \cdot g\left(\frac{1}{n}\right) \\
=\left[g\binom{1}{n}\right]^{\mathrm{m}} .
\end{array}
$$

In particular, let $m=n$ then
$$
\left[g\left(\frac{1}{n}\right)\right]^{n}=g(1)=a_{0}
$$

Combining (6), we get
$$
g\left(\frac{1}{n}\right)=a^{1} .
$$

From (7), we have
$$
g\left(\frac{m}{n}\right)=\left[g\left(\frac{1}{n}\right)\right]^{m}=a^{\frac{m}{2}} .
$$

Finally, in (5), let $x_{1}=-x_{2}=x$, we get
$$
g(x) \cdot g(-x)=g(0)=1,
$$

thus, $g(-x)=\frac{1}{g(x)}$.
Therefore, from ( ) ( ) we get
$$
g\left(-\frac{m}{n}\right)=\frac{1}{g\left(\frac{m}{n}\right)}=a^{-\frac{m}{n}} .
$$

Thus, we have proved that for any rational number $x$, we have
$$
g(x)=a^{x},
$$

so, $f(x)=\omega \cdot g(x)=\omega \cdot a^{x}$.
Here $\omega^{1987}=1, a>0$.
In summary, all such functions are
$$
f(x)=\omega \cdot a^{x},
$$

where the complex number $\omega$ satisfies $\omega^{1987}=\omega$, and the real number $a>0$.