(i) To find stationary points, we set the derivative equal to zero. The derivative is \(\frac{dy}{dx} = 3x^2 - 4x + 5\). The discriminant of this quadratic is \(b^2 - 4ac = (-4)^2 - 4 \times 3 \times 5 = 16 - 60 = -44\), which is negative. Therefore, there are no real roots, and the curve has no stationary points.

(ii) The gradient \(m = 3x^2 - 4x + 5\). To find the stationary value of \(m\), we differentiate \(m\) with respect to \(x\): \(\frac{dm}{dx} = 6x - 4\). Setting \(\frac{dm}{dx} = 0\) gives \(6x - 4 = 0\), so \(x = \frac{2}{3}\). Substituting \(x = \frac{2}{3}\) into \(m = 3x^2 - 4x + 5\), we get \(m = 3 \left( \frac{2}{3} \right)^2 - 4 \left( \frac{2}{3} \right) + 5 = \frac{11}{3}\).

To determine the nature, we find \(\frac{d^2m}{dx^2} = 6\), which is positive, indicating a minimum.