Answer
$-\frac{\sqrt {26}}{26}$
Work Step by Step
Step 1. Given $sin\alpha=\frac{3}{5}, \frac{\pi}{2}\lt\alpha\lt\pi$, we know $\alpha$ is in quadrant II; thus $cos\alpha=-\sqrt {1-sin^2\alpha}=-\frac{4}{5}$ and $tan\alpha=-\frac{3}{4}$
Step 2. Given $cos\beta=-\frac{12}{13}, \pi\lt\beta\lt \frac{3\pi}{2}$, we know $\beta$ is in quadrant III; thus $sin\beta=-\sqrt {1-cos^2\beta}=-\frac{5}{13}$ and $tan\beta=\frac{5}{12}$
Step 3. Use the Half-Angle Formula, with $\frac{\pi}{2}\lt \frac{\beta}{2} \lt \frac{3\pi}{4}$, we have
$cos( \frac{\beta}{2})=-\sqrt {\frac{1+cos\beta}{2}}=-\sqrt {\frac{1+(-\frac{12}{13})}{2}}=-\frac{\sqrt {26}}{26}$
