解:原式$=(15a^2b - 5ab^2)-(-4ab^2+12a^2b)-2a^2b $
$= 15a^2b-5ab^2+4ab^2-12a^2b - 2a^2b$
$=a^2b - ab^2.$
$ $当$a =-\frac {1}{2},$$b = 2$时,
$ $原式$=(-\frac {1}{2})^2×2-(-\frac {1}{2})×2^2=\frac {1}{4}×2+\frac {1}{2}×4=\frac {1}{2}+2=\frac {5}{2}$