Given \(\frac{dy}{dx} = 3x^2 + ax + b\), stationary points occur where \(\frac{dy}{dx} = 0\).

For \(x = -1\), \(3(-1)^2 + a(-1) + b = 0\) gives:

\(3 - a + b = 0\)

For \(x = 3\), \(3(3)^2 + a(3) + b = 0\) gives:

\(27 + 3a + b = 0\)

Solving these equations simultaneously:

1. \(3 - a + b = 0\)

2. \(27 + 3a + b = 0\)

Subtract equation 1 from equation 2:

\(27 + 3a + b - (3 - a + b) = 0\)

\(27 + 3a + b - 3 + a - b = 0\)

\(24 + 4a = 0\)

\(a = -6\)

Substitute \(a = -6\) into equation 1:

\(3 - (-6) + b = 0\)

\(3 + 6 + b = 0\)

\(b = -9\)

Integrate \(\frac{dy}{dx} = 3x^2 - 6x - 9\) to find \(y\):

\(y = x^3 - 3x^2 - 9x + c\)

Use point \((-1, 2)\):

\(2 = (-1)^3 - 3(-1)^2 - 9(-1) + c\)

\(2 = -1 - 3 + 9 + c\)

\(c = -3\)

Use point \((3, k)\):

\(k = 3^3 - 3(3)^2 - 9(3) - 3\)

\(k = 27 - 27 - 27 - 3\)

\(k = -30\)

(i) Integrate \(\frac{d^2y}{dx^2} = 2x - 5\) to find \(\frac{dy}{dx} = x^2 - 5x + c\).

Since the curve has a stationary point at \(x = 3\), \(\frac{dy}{dx} = 0\) when \(x = 3\).

Substitute \(x = 3\) into \(x^2 - 5x + c = 0\):

\(9 - 15 + c = 0\)

\(c = 6\)

Integrate again to find \(y = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + d\).

Use the point (3, 6) to find \(d\):

\(6 = \frac{27}{3} - \frac{45}{2} + 18 + d\)

\(6 = 9 - 22.5 + 18 + d\)

\(d = \frac{1}{2}\)

Thus, the equation of the curve is \(y = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + \frac{1}{2}\).

(ii) To find the other stationary point, set \(\frac{dy}{dx} = 0\):

\(x^2 - 5x + 6 = 0\)

Factorize to get \((x - 2)(x - 3) = 0\).

The other stationary point is at \(x = 2\).

(iii) Determine the nature of the stationary points using \(\frac{d^2y}{dx^2} = 2x - 5\):

At \(x = 3\), \(\frac{d^2y}{dx^2} = 2(3) - 5 = 1\) (positive, so minimum).

At \(x = 2\), \(\frac{d^2y}{dx^2} = 2(2) - 5 = -1\) (negative, so maximum).

To find \(f(x)\), integrate \(f'(x) = kx^2 - 2x\):

\(f(x) = \frac{k}{3}x^3 - x^2 + c\)

Use the point \((0, 2)\):

\(2 = \frac{k}{3}(0)^3 - (0)^2 + c\)

\(c = 2\)

Use the point \((3, -1)\):

\(-1 = \frac{k}{3}(3)^3 - (3)^2 + 2\)

\(-1 = 9k - 9 + 2\)

\(-1 = 9k - 7\)

\(9k = 6\)

\(k = \frac{2}{3}\)