Answer
(a) $\frac{1}{2}$
(b) $-\frac{1+\sqrt 3}{2}$
Work Step by Step
$sin(u-v)=sin~u~cos~v-cos~u~sin~v$
(a) $sin(\frac{7\pi}{6}-\frac{\pi}{3})=sin~\frac{7\pi}{6}~cos~\frac{\pi}{3}-cos~\frac{7\pi}{6}~sin~\frac{\pi}{3}=-\frac{1}{2}~\frac{1}{2}-(-\frac{\sqrt 3}{2})\frac{\sqrt 3}{2}=-\frac{1}{4}+\frac{3}{4}=\frac{2}{4}=\frac{1}{2}$
(b) $sin~\frac{7\pi}{6}-sin~\frac{\pi}{3}=-\frac{1}{2}-\frac{\sqrt 3}{2}=-\frac{1+\sqrt 3}{2}$
