(i) The area of sector AOB is \(\frac{1}{2}r^2\theta\).

In \(\triangle OAX\), \(AX = r\sin\theta\) and \(OX = r\cos\theta\).

The area of \(\triangle OAX\) is \(\frac{1}{2} \times OX \times AX = \frac{1}{2}r^2\sin\theta\cos\theta\).

Thus, the area of the shaded region AXB is:

\(A = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\cos\theta = \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)\)

(ii) Given \(r = 12\) and \(\theta = \frac{1}{6}\pi\):

\(AX = 12\sin\frac{1}{6}\pi = 6\)

\(OX = 12\cos\frac{1}{6}\pi = 6\sqrt{3}\)

\(BX = OB - OX = 12 - 6\sqrt{3}\)

The length of arc AB is \(r\theta = 12 \times \frac{1}{6}\pi = 2\pi\).

Therefore, the perimeter of the shaded region AXB is:

\(P = AX + BX + \text{arc } AB = 6 + (12 - 6\sqrt{3}) + 2\pi = 18 - 6\sqrt{3} + 2\pi\)