2. $f(x)$ is a function defined on the set of non-negative real numbers, and takes non-negative real values. Find all functions $f(x)$ that satisfy the following conditions:
(1) $f[x f(y)] f(y)=f(x+y)$;
(2) $f(2)=0$;
(3) For $0 \leqslant x<2$, $f(x) \neq 0$.
2. In (1) let $y=2, x \geqslant 0$, then using (2) we have $f(x+2)=0$. This shows that when $x \geqslant 2$, $f(x) \equiv 0$.

When $0 \leqslant x<2$, from conditions (1), (2), and (3) we get $f[(2-x) f(x)]=0$. But by (3) and $x \geqslant 2$ when $f(x) \equiv 0$, we know $f(x) \geqslant \frac{2}{2-x}$.

On the other hand, $f\left(\frac{2}{f(x)}+x\right)=f\left[\frac{2}{f(x)} \cdot f(x)\right] \cdot f(x)=f(2) f(x)=0$. Therefore, $\frac{2}{f(x)}+x \geqslant 2$, which means $f(x) \leqslant \frac{2}{2-x}$.
Thus, $f(x)=\left\{\begin{array}{ll}\frac{2}{2-x} & x \in[0,2) \\ 0 & x \in[2,+\infty)\end{array}\right.$
It is easy to verify: $f(x)$ is the desired function.