(i) Given \(\frac{dy}{dx} = 6x^2 + \frac{k}{x^3}\) and the gradient at \(P(1, 9)\) is 2, substitute \(x = 1\) into the derivative:

\(6(1)^2 + \frac{k}{(1)^3} = 2\)

\(6 + k = 2\)

Solving for \(k\), we get \(k = -4\).

(ii) To find the equation of the curve, integrate \(\frac{dy}{dx} = 6x^2 - \frac{4}{x^3}\):

\(y = \int (6x^2 - \frac{4}{x^3}) \, dx\)

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

Using the point \(P(1, 9)\), substitute \(x = 1\) and \(y = 9\):

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

\(9 = 2 + 2 + c\)

\(c = 5\)

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

(a) To find \(f(x)\), integrate \(f'(x) = 2x^2 - 7 - \frac{4}{x^2}\):

\(f(x) = \int (2x^2 - 7 - \frac{4}{x^2}) \, dx = \frac{2}{3}x^3 - 7x + 4x^{-1} + c\).

Given \(f(1) = -\frac{1}{3}\), substitute to find \(c\):

\(-\frac{1}{3} = \frac{2}{3}(1)^3 - 7(1) + 4(1)^{-1} + c\)

\(-\frac{1}{3} = \frac{2}{3} - 7 + 4 + c\)

\(-\frac{1}{3} = -\frac{1}{3} + c\)

\(c = 2\)

Thus, \(f(x) = \frac{2}{3}x^3 - 7x + 4x^{-1} + 2\).

(b) Stationary points occur where \(f'(x) = 0\):

\(2x^2 - 7 - \frac{4}{x^2} = 0\)

\(2x^4 - 7x^2 - 4 = 0\)

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

\(x^2 = 4 \Rightarrow x = \pm 2\)

For \(x = 2\), \(f(2) = \frac{2}{3}(2)^3 - 7(2) + 4(2)^{-1} + 2 = \frac{14}{3}\)

For \(x = -2\), \(f(-2) = \frac{2}{3}(-2)^3 - 7(-2) + 4(-2)^{-1} + 2 = -\frac{26}{3}\)

Coordinates: \((2, \frac{14}{3})\) and \((-2, -\frac{26}{3})\).

(c) Differentiate \(f'(x)\) to find \(f''(x)\):

\(f''(x) = \frac{d}{dx}(2x^2 - 7 - \frac{4}{x^2}) = 4x + 8x^{-3}\).

(d) Evaluate \(f''(x)\) at the stationary points:

\(f''(2) = 4(2) + 8(2)^{-3} = 9 > 0\), so minimum at \(x = 2\).

\(f''(-2) = 4(-2) + 8(-2)^{-3} = -9 < 0\), so maximum at \(x = -2\).

To find the equation of the curve, we need to integrate the given derivative \(\frac{dy}{dx} = 2 - 8(3x + 4)^{-\frac{1}{2}}\).

Integrate term by term:

\(\int 2 \, dx = 2x + C_1\)

For the second term, let \(u = 3x + 4\), then \(du = 3 \, dx\) or \(dx = \frac{du}{3}\).

\(\int -8(3x + 4)^{-\frac{1}{2}} \, dx = -8 \int u^{-\frac{1}{2}} \cdot \frac{du}{3}\)

\(= -\frac{8}{3} \int u^{-\frac{1}{2}} \, du\)

\(= -\frac{8}{3} \cdot 2u^{\frac{1}{2}} + C_2\)

\(= -\frac{16}{3} \sqrt{3x+4} + C_2\)

Combine the results:

\(y = 2x - \frac{16}{3} \sqrt{3x+4} + C\)

Given that the curve intersects the y-axis where \(y = \frac{4}{3}\), substitute \(x = 0\) and \(y = \frac{4}{3}\) to find \(C\):

\(\frac{4}{3} = 2(0) - \frac{16}{3} \sqrt{4} + C\)

\(\frac{4}{3} = -\frac{16}{3} \cdot 2 + C\)

\(\frac{4}{3} = -\frac{32}{3} + C\)

\(C = \frac{4}{3} + \frac{32}{3} = 12\)

Thus, the equation of the curve is:

\(y = 2x + \frac{8}{3}\sqrt{3x+4} + 12\)

(i) At \(x = 4\), \(\frac{dy}{dx} = 2\).

Using the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = 2 \times 3 = 6\).

(ii) Integrate \(\frac{dy}{dx} = 1 + 2x^{-\frac{1}{2}}\) to find \(y\).

\(y = x + 4x^{\frac{1}{2}} + c\).

Substitute \(x = 4, y = 6\):

\(6 = 4 + 4(4^{\frac{1}{2}}) + c\)

\(c = -6\)

Thus, \(y = x + 4x^{\frac{1}{2}} - 6\).

(iii) Equation of the tangent at A is \(y - 6 = 2(x - 4)\) or \(y = 2x - 8\).

The tangent crosses the x-axis at \(B = (1, 0)\).

The gradient of the normal is \(-\frac{1}{2}\), so the equation is \(y = -\frac{1}{2}x + 8\).

The normal crosses the x-axis at \(C = (16, 0)\).

The area of triangle ABC is \(\frac{1}{2} \times 15 \times 6 = 45\).

(i) Given \(f''(x) = \frac{12}{x^3}\), integrate to find \(f'(x)\):

\(f'(x) = \int \frac{12}{x^3} \, dx = -\frac{6}{x^2} + c\).

Since \(f'(2) = 0\), substitute \(x = 2\):

\(0 = -\frac{6}{4} + c \Rightarrow c = \frac{3}{2}\).

Thus, \(f'(x) = -\frac{6}{x^2} + \frac{3}{2}\).

Integrate \(f'(x)\) to find \(f(x)\):

\(f(x) = \int \left(-\frac{6}{x^2} + \frac{3}{2}\right) \, dx = \frac{6}{x} + \frac{3x}{2} + A\).

Since \(f(2) = 10\), substitute \(x = 2\):

\(10 = \frac{6}{2} + 3 + A \Rightarrow A = 4\).

Thus, \(f(x) = \frac{6}{x} + \frac{3x}{2} + 4\).

(ii) Set \(f'(x) = 0\):

\(-\frac{6}{x^2} + \frac{3}{2} = 0 \Rightarrow x = \pm 2\).

The other stationary point is \((-2, -2)\).

(iii) Evaluate \(f''(x)\) at the stationary points:

At \(x = 2\), \(f''(2) = \frac{12}{8} = 1.5\), indicating a minimum.

At \(x = -2\), \(f''(-2) = -1.5\), indicating a maximum.