(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 3x^2 - 4x + 1\).

\(y = \int (3x^2 - 4x + 1) \, dx = x^3 - 2x^2 + x + c\).

Use the point (1, 5) to find \(c\):

\(5 = 1^3 - 2(1)^2 + 1 + c\)

\(5 = 1 - 2 + 1 + c\)

\(5 = 0 + c\)

\(c = 5\)

Thus, the equation of the curve is \(y = x^3 - 2x^2 + x + 5\).

(ii) To find where the gradient is positive, solve \(3x^2 - 4x + 1 > 0\).

Factorize or use the quadratic formula to find the roots:

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

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}\)

\(x = \frac{4 \pm \sqrt{16 - 12}}{6}\)

\(x = \frac{4 \pm 2}{6}\)

\(x = 1\) or \(x = \frac{1}{3}\)

The inequality \(3x^2 - 4x + 1 > 0\) holds for \(x < \frac{1}{3}\) and \(x > 1\).