Let \(a=\sec^{-1}x\) and \(b=\text{cosec}^{-1}x\). For the principal branches, \(a\in[0,\pi]\setminus\{\pi/2\}\) and it is a standard identity that \(a+b=\dfrac{\pi}{2}\), so \(b=\dfrac{\pi}{2}-a\).

Let \(g(a)=a^2+b^2=a^2+\left(\dfrac{\pi}{2}-a\right)^2=2a^2-\pi a+\dfrac{\pi^2}{4}\).

This is a upward-opening parabola in \(a\) with vertex (minimum) at \(a=\dfrac{\pi}{4}\), which lies in the allowed domain \([0,\pi]\setminus\{\pi/2\}\).

Minimum value: \(g\left(\dfrac{\pi}{4}\right)=2\cdot\dfrac{\pi^2}{16}-\dfrac{\pi^2}{4}+\dfrac{\pi^2}{4}=\dfrac{\pi^2}{8}\).

Since the parabola opens upward, its maximum over \([0,\pi]\) occurs at an endpoint. Compare \(g(0)=\dfrac{\pi^2}{4}\) with \(g(\pi)=2\pi^2-\pi^2+\dfrac{\pi^2}{4}=\dfrac{5\pi^2}{4}\).

Since \(\dfrac{5\pi^2}{4}>\dfrac{\pi^2}{4}\), the maximum value is \(g(\pi)=\dfrac{5\pi^2}{4}\).

Sum of maximum and minimum of \(a^2+b^2\): \(\dfrac{5\pi^2}{4}+\dfrac{\pi^2}{8}=\dfrac{10\pi^2+\pi^2}{8}=\dfrac{11\pi^2}{8}\).

Multiplying by 16: \(16\times\dfrac{11\pi^2}{8}=22\pi^2\).