(i) To find the equation of line \(AB\), calculate the gradient: \(\text{Gradient of } AB = \frac{2 - 0}{5 - 1} = \frac{1}{2}\).

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

(ii) The derivative of the curve \(y = (x - 1)^{\frac{1}{2}}\) is \(\frac{dy}{dx} = \frac{1}{2}(x - 1)^{-\frac{1}{2}}\).

Set \(\frac{dy}{dx} = \frac{1}{2}\) to find the tangent parallel to \(AB\):

\(\frac{1}{2}(x - 1)^{-\frac{1}{2}} = \frac{1}{2}\) gives \(x = 2\), \(y = 1\).

The equation of the tangent is \(y - 1 = \frac{1}{2}(x - 2)\), simplifying to \(y = \frac{1}{2}x\).

(iii) The perpendicular distance \(d\) between the line \(AB\) and the tangent is given by:

\(d = \sin(26.5^\circ) = 0.45\).