10 A function $f(x)$ has a domain $D$. If for any $x_{1}, x_{2} \in D$, when $x_{1}<x_{2}$, we have $f\left(x_{1}\right) \leqslant f\left(x_{2}\right)$, then the function $f(x)$ is called non-decreasing on $D$. Suppose the function $f(x)$ is non-decreasing on $[0,1]$ and satisfies the following three conditions: (1) $f(0)=0$; (2) $f\left(\frac{x}{3}\right)=\frac{1}{2} f(x)$; (3) $f(1-x)=1-f(x)$. Then $f\left(\frac{5}{12}\right)+f\left(\frac{1}{8}\right)=$ $\qquad$
Taking $x=\frac{1}{2}$ in (2), we get $f\left(\frac{1}{6}\right)=\frac{1}{2} f\left(\frac{1}{2}\right)=\frac{1}{4}$;
Taking $x=\frac{1}{3}$ in (2), we get $f\left(\frac{1}{9}\right)=\frac{1}{2} f\left(\frac{1}{3}\right)=\frac{1}{4}$.
Since the function is non-decreasing, $\frac{1}{9}<\frac{1}{8}<\frac{1}{6}$, hence $f\left(\frac{1}{8}\right)=\frac{1}{4}$, so
$$
f\left(\frac{5}{12}\right)+f\left(\frac{1}{8}\right)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4} .
$$