(i) The line cuts the x-axis at A, so the y-coordinate of A is 0. The angle \(\angle BAO = \frac{1}{6} \pi\) radians implies that the slope of the line is \(\tan\left(\frac{1}{6} \pi\right) = \frac{1}{\sqrt{3}}\).

The equation of the line is \(y - 2 = \frac{1}{\sqrt{3}}(x - 0)\).

Setting \(y = 0\) to find the x-coordinate of A:

\(0 - 2 = \frac{1}{\sqrt{3}}x\)

\(x = 2\sqrt{3}\)

(ii) The midpoint of AB is \(\left(\frac{\sqrt{3}}{2}, 1\right)\).

The gradient of AB is \(-\frac{1}{\sqrt{3}}\), so the gradient of the perpendicular bisector is \(\sqrt{3}\).

The equation of the perpendicular bisector is \(y - 1 = \sqrt{3}(x - \frac{\sqrt{3}}{2})\).

Simplifying gives \(y = \sqrt{3}x - 2\).