2. Let $f$ be a function defined on the set of non-negative real numbers and taking values in the same set. Find all functions $f$ that satisfy the following conditions:
(i) $f[x \cdot f(y)] \cdot f(y)=f(x+y)$;(ii) $f(2)=0$; (iii) $f(x) \neq 0$, when $0 \leqslant x<2$.
2. Let $f$ satisfy the known conditions. When $x \geqslant 2$, we have $f(x)=f[(x-2)+2]=f[(x-2) \cdot f(2)] \cdot f(2)=0$, so $f(x)=0$. If $0 \leqslant x<2$, let $y \geqslant 2-x$, then $x+y \geqslant 2, f(x+y)=0, f[y f(x)] \cdot f(x)=f(y+x)=$ $f(x+y)=0$. Also, by the known condition $f(x) \neq 0$ when $0 \leqslant x<2$, thus $f(y f(x))=0, y f(x) \geqslant 2, y \geqslant \frac{2}{f(x)}$. This means that from $y \geqslant 2-x$, we can get $y \geqslant \frac{2}{f(x)}$. But by the reversibility of the above reasoning process, we know that from $y \geqslant \frac{2}{f(x)}$ we can also get $y \geqslant 2-x$, therefore, we have $2-x=\frac{2}{f(x)}, f(x)=\frac{2}{2-x}(0 \leqslant x<2$ when $)$, hence the function that satisfies the conditions is
$$
f(x)=\left\{\begin{array}{l}
\frac{2}{2-x}, 0 \leqslant x<2 \text { when, } \\
0, \text { when } x \geqslant 2 \text { . }
\end{array}\right.
$$