(i) Differentiate \(y = \frac{1}{6}(2x - 3)^3 - 4x\) with respect to \(x\).

Using the chain rule, \(\frac{dy}{dx} = \frac{1}{6} \times 3 \times (2x - 3)^2 \times 2 - 4\).

(ii) To find the y-intercept, set \(x = 0\):

\(y = \frac{1}{6}(2 \times 0 - 3)^3 - 4 \times 0 = -\frac{27}{6} = -\frac{9}{2}\).

The slope of the tangent is \(\frac{dy}{dx}\) at \(x = 0\):

\(\frac{dy}{dx} = \frac{1}{6} \times 3 \times (-3)^2 \times 2 - 4 = 5\).

The equation of the tangent is \(y + \frac{9}{2} = 5x\).

Rearranging gives \(2y + 9 = 10x\).

(iii) The function is increasing when \(\frac{dy}{dx} > 0\).

\(\frac{1}{6} \times 3 \times (2x - 3)^2 \times 2 - 4 > 0\).

\((2x - 3)^2 > 4\).

Solving \((2x - 3)^2 - 4 = 0\) gives \(x = \frac{5}{2}\) or \(x = \frac{1}{2}\).

The set of values for which the function is increasing is \(x > \frac{5}{2}, x < \frac{1}{2}\).