(i) The minimum point occurs when \(y = 0\). Solving \(x(x-2)^2 = 0\) gives \(x = 0\) or \(x = 2\). Thus, \(a = 2\).

(ii) To find \(b\), differentiate \(y = x(x-2)^2\) to get \(\frac{dy}{dx} = 3x^2 - 8x + 4\). Setting \(\frac{dy}{dx} = 0\) gives \((x-2)(3x-2) = 0\), so \(x = 2\) or \(x = \frac{2}{3}\). The maximum point is at \(x = \frac{2}{3}\), so \(b = \frac{2}{3}\).

(iii) The area of the shaded region is \(\int_0^2 x(x-2)^2 \, dx\). Expanding gives \(x^3 - 4x^2 + 4x\). Integrating, we get \(\left[ \frac{x^4}{4} - \frac{4x^3}{3} + 2x^2 \right]_0^2 = \frac{4}{3}\).

(iv) The second derivative is \(\frac{d^2y}{dx^2} = 6x - 8\). Setting \(\frac{d^2y}{dx^2} = 0\) gives \(x = \frac{4}{3}\). At this point, \(\frac{dy}{dx} = -\frac{4}{3}\), so \(m = -\frac{4}{3}\).