(i) To find where the gradient is less than 5, first find the derivative: \(f'(x) = 3x^2 - 4x + 1\).

Set \(f'(x) < 5\):

\(3x^2 - 4x + 1 < 5\)

\(3x^2 - 4x - 4 < 0\)

Factorize: \((3x + 2)(x - 2) < 0\)

Find the critical points: \(x = -\frac{2}{3}\) and \(x = 2\).

The solution is \(-\frac{2}{3} < x < 2\).

(ii) To find stationary points, set \(f'(x) = 0\):

\(3x^2 - 4x + 1 = 0\)

Factorize: \((3x - 1)(x - 1) = 0\)

Solutions: \(x = \frac{1}{3}\) and \(x = 1\).

Find \(f(x)\) at these points:

\(f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - 2\left(\frac{1}{3}\right)^2 + \frac{1}{3} = \frac{4}{27}\)

\(f(1) = 1^3 - 2 \times 1^2 + 1 = 0\)

Determine nature using second derivative: \(f''(x) = 6x - 4\).

\(f''\left(\frac{1}{3}\right) = 6\left(\frac{1}{3}\right) - 4 = -2\) (negative, so maximum)

\(f''(1) = 6 \times 1 - 4 = 2\) (positive, so minimum)

Maximum at \(\left( \frac{1}{3}, \frac{4}{27} \right)\); Minimum at \((1, 0)\).