(5) Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies:
(1) $f(1)=1$; (2) when $00$; (3) for any real numbers $x, y$, $f(x+y)-f(x-y)=2 f(1-x) f(y)$. Then $f\left(\frac{1}{3}\right)=$ . $\qquad$
(5) $\frac{1}{2}$ Hint: In the conditional equation (3), let $x=\frac{2}{3}, y=\frac{1}{3}$, we get
$$
f(1)-f\left(\frac{1}{3}\right)=2 f\left(\frac{1}{3}\right) f\left(\frac{1}{3}\right) \text {, }
$$

Combining $f(1)=1, f\left(\frac{1}{3}\right)>0$, we solve to get $f\left(\frac{1}{3}\right)=\frac{1}{2}$.