解:原式$=\frac{2}{3}ab^{2}+\frac{2}{3}a^{2}b - 4,$当$a = -\frac{1}{2},$$b = 3$时,
将$a = -\frac{1}{2},$$b = 3$代入$\frac{2}{3}ab^{2}+\frac{2}{3}a^{2}b - 4$得:
$\frac{2}{3}×(-\frac{1}{2})×3^{2}+\frac{2}{3}×(-\frac{1}{2})^{2}×3 - 4$
$=\frac{2}{3}×(-\frac{1}{2})×9+\frac{2}{3}×\frac{1}{4}×3 - 4$
$=-3+\frac{1}{2}-4$
$=-\frac{6}{2}+\frac{1}{2}-\frac{8}{2}$
$=-\frac{13}{2}$