Critical numbers of tan(x)

The derivative is $f'(x) = \sec^2(x) = \dfrac{1}{\cos^2(x)}$. Since $\cos(x) \ne 0$ on $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, $f'(x)$ is always defined on this open interval.

Also, $\sec^2(x) \ge 1 > 0$ for all $x$ in this interval, so $f'(x)$ is never zero. Therefore there are no points in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ where $f'(x) = 0$ or $f'(x)$ is undefined.

The Extreme Value Theorem requires $f$ to be continuous on a closed interval $[a,b]$. Here the interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ is open, not closed, so the theorem's hypothesis is not satisfied and it gives no guarantee of extrema.

As $x \to \left(\dfrac{\pi}{2}\right)^-$, $\tan(x) \to +\infty$, and as $x \to \left(-\dfrac{\pi}{2}\right)^+$, $\tan(x) \to -\infty$. Since $f$ increases without bound near the right endpoint and decreases without bound near the left endpoint, and the interval has no critical numbers, there is no absolute maximum or absolute minimum on this interval.

Yes. On a closed interval, the Extreme Value Theorem guarantees absolute extrema exist regardless of critical numbers, because they can occur at the endpoints. Even if a continuous function has no interior critical numbers, its absolute maximum and minimum simply occur at the two endpoints. This contrasts with the open interval case for tangent, where there are no endpoints to fall back on.