解:$(1)$当$x=\frac 12$时,原式$=\frac {5×\frac 12}{3×(\frac 12)^2-2}=-2$
$(2)$当$a=\frac {5}{6}$,$b=\frac {1}{6}$时,$a+b=\frac {5}{6}+\frac {1}{6}=1$,$a-b=\frac {5}{6}- \frac {1}{6}=\frac {2}{3}$
$a-2b=\frac {5}{6}-\frac {2}{6}=\frac {1}{2}$
∴原式$=\frac {(a+b)(a-b)}{3(a-2b)}=\frac {1×\frac 23}{3×\frac 12}=\frac {4}{9}$