Answer
The absolute maximum $ f(\pi/4)=1$ and the absolute minimum $f(-\pi/3)=-\sqrt 3$.
See graph.
Work Step by Step
Step 1. Given the function $f(\theta)=tan\theta, -\pi/3\leq \theta\leq \pi/4$, we have $f(-\pi/3)=-\sqrt 3, f(\pi/4)=1$,
Step 2. $f'(\theta)=sec^2\theta\ne0$, no critical points.
Step 3. We can identify the absolute maximum as $ f(\pi/4)=1$ and the absolute minimum as $f(-\pi/3)=-\sqrt 3$ on the given interval.
Step 4. See graph.