(i) Differentiate the curve \(y = (6x + 2)^{\frac{1}{3}}\) to find the gradient at \(A\):

\(\frac{dy}{dx} = 6 \times \frac{1}{3}(6x + 2)^{-\frac{2}{3}} = 2(6x + 2)^{-\frac{2}{3}}\)

At \(x = 1\), \(\frac{dy}{dx} = 2(8)^{-\frac{2}{3}} = \frac{1}{2}\).

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

The gradient of the normal is \(-2\) (negative reciprocal of \(\frac{1}{2}\)).

The equation of the normal is \(y - 2 = -2(x - 1)\).

(ii) The coordinates of \(B\) are \((0, \frac{1}{2})\) and \(C\) are \((2, 0)\).

Distance \(BC = \sqrt{(2 - 0)^2 + (0 - \frac{1}{2})^2} = \frac{5}{2}\).

(iii) The equation of \(BC\) is \(y - \frac{1}{2} = -\frac{3}{4}(x - 0)\) or \(y = -\frac{3}{4}x + \frac{1}{2}\).

Intersection \(E\) is found by solving \(-\frac{3}{4}x + \frac{1}{2} = 2x\).

\(x = \frac{6}{11}, \; y = \frac{12}{11}\).

The mid-point of \(OA\) is \(\left( \frac{1}{2}, 1 \right)\), so \(E\) is not the mid-point.