48. Let the function $y=f(x)$ be a decreasing function defined on $\mathrm{R}_{+}$, and it satisfies $f(x y)=f(x)+f(y), f\left(\frac{1}{3}\right)=1$,
(1) Find the value of $f(1)$;
(2) If $f(x)+f(2-x)<2$, find the range of $x$.
\(\begin{array}{l}\text { 48. }(1) \because f(1)=f(1)+f(1), \\ \therefore f(1)=0 . \\ (2) \because f(x)+f(2-x)=f(x(2-x)), \\ \therefore f(x(2-x))=0, \\ 2-x>0, \\ x(2-x)>\frac{1}{9},\end{array} \text { solving this yields } \quad 0<x<1+\frac{\sqrt{10}}{3} .\right.\end{array}\)