(i) Since AX is tangent to the circle at A, \(\angle OAX = \frac{\pi}{2}\). Using the tangent formula, \(AX = 6 \tan \frac{\pi}{3} = 6 \sqrt{3}\).

(ii) The area of triangle OAX is \(\frac{1}{2} \times 6 \times 6 \sqrt{3} = 18 \sqrt{3}\).

The area of sector OAB is \(\frac{1}{2} \times 6^2 \times \frac{\pi}{3} = 6\pi\).

Thus, the area of the shaded region is \(18 \sqrt{3} - 6\pi\).

(iii) The arc length AB is \(6 \times \frac{\pi}{3} = 2\pi\).

Using trigonometry, \(OX = \frac{6}{\cos \frac{\pi}{3}} = 12\), so \(BX = 6\).

The perimeter of the shaded region is \(6 \sqrt{3} + 2\pi + 6\).