Answer
$\frac{56}{33}$
Work Step by Step
Step 1. Given $sin\alpha=\frac{4}{5}, \frac{\pi}{2}\lt\alpha\lt\pi$, we can identify $\alpha$ to be in quadrant II, which means
$cos\alpha=-\sqrt {1-sin^2\alpha}=-\frac{3}{5}$ and $tan\alpha=-\frac{4}{3}$
Step 2. Given $cos\beta=\frac{5}{13}, 0\lt\beta\lt \frac{\pi}{2}$, we can identify $\beta$ to be in quadrant I, which means
$sin\beta=\sqrt {1-cos^2\beta}=\frac{12}{13}$ and $tan\beta=\frac{12}{5}$
Step 3. Use the the subtraction formula, we have
$tan(\alpha-\beta)=\frac{tan\alpha-tan\beta}{1+tan\alpha tan\beta}=\frac{(-\frac{4}{3})-(\frac{12}{5})}{1+(-\frac{4}{3})(\frac{12}{5})}=\frac{56}{33}$
