(a)(i) Since \(AX\) is a tangent, \(\angle BAO = \angle OBA = \frac{\pi}{2} - \alpha\). Therefore, \(\angle AOB = \pi - (\frac{\pi}{2} - \alpha) - (\frac{\pi}{2} - \alpha) = 2\alpha\).

(a)(ii) The area of the sector \(OAB\) is \(\frac{1}{2} r^2 (2\alpha)\). The area of triangle \(OAB\) is \(\frac{1}{2} r^2 \sin 2\alpha\). Thus, the area of the shaded segment is \(\frac{1}{2} r^2 (2\alpha) - \frac{1}{2} r^2 \sin 2\alpha\).

(b) Use \(\alpha = \frac{\pi}{6}\) and \(r = 4\). The area of one segment is \(\frac{1}{2} (4)^2 \left( \frac{\pi}{3} \right) - \frac{1}{2} (4)^2 \sin \frac{\pi}{3} = \frac{8\pi}{3} - 4\sqrt{3}\).

The area of triangle \(ABC\) is \(\frac{1}{2} (4)^2 \sin \frac{\pi}{3} = 4\sqrt{3}\).

The area of the shaded region is \(3 \left( \frac{8\pi}{3} - 4\sqrt{3} \right) - 4\sqrt{3} = 16\sqrt{3} - 8\pi\).