Example 9 Let $f(x)$ be defined for $x>0$, $f(2)=1$, $f(xy)=f(x)+f(y)$, and the necessary and sufficient condition for $f(x)>f(y)$ to hold is $x>y>0$. Find: (1) the values of $f(1)$ and $f(4)$; (2) the range of values for $x$ such that $f(x)+f(x-3) \leqslant 2$.

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(1) Since $f(2)=1$, and for $x>0, y>0, f(x y)=f(x)+f(y)$, let $x=1, y=2$. We get $f(2)=f(1)+f(2)$, which means $f(1)=0$.
Let $x=2, y=2$, we get $f(4)=f(2)+f(2)=2$.
(2) From the condition $f(x y)=f(x)+f(y)$, we have
$$
f(x)+f(x-3)=f\left(x^{2}-3 x\right) .
$$

Also, since $f(4)=2$, from $f(x)+f(x-3) \leqslant 2$, we get
$$
f\left(x^{2}-3 x\right) \leqslant f(4) \text {. }
$$

By the condition that $f(x)>f(y)$ holds if and only if $x>y>0$, we have $\left\{\begin{array}{l}x^{2}-3 x \leqslant 4 \text {. } \\ x>0, \\ x-3>0 .\end{array}\right.$ Solving this, we get $3<x \leqslant 4$.