(i) To find the equation of the normal, first find the derivative \(\frac{dy}{dx}\) of the curve \(y = \frac{4}{5 - 3x}\).

\(\frac{dy}{dx} = \frac{-4}{(5 - 3x)^2} \times (-3) = \frac{12}{(5 - 3x)^2}\).

At \(x = 1\), \(\frac{dy}{dx} = \frac{12}{(5 - 3 \times 1)^2} = \frac{12}{4} = 3\).

The gradient of the tangent is 3, so the gradient of the normal is \(-\frac{1}{3}\).

The point on the curve at \(x = 1\) is \((1, 2)\).

The equation of the normal is \(y - 2 = -\frac{1}{3}(x - 1)\).

Simplifying gives \(y = -\frac{1}{3}x + \frac{7}{3}\).

(ii) The volume of the solid of revolution is given by \(V = \pi \int_{0}^{1} y^2 \, dx\).

\(V = \pi \int_{0}^{1} \left( \frac{4}{5 - 3x} \right)^2 \, dx = \pi \int_{0}^{1} \frac{16}{(5 - 3x)^2} \, dx\).

Let \(u = 5 - 3x\), then \(du = -3 \, dx\) or \(dx = -\frac{1}{3} \, du\).

Change the limits: when \(x = 0, u = 5\); when \(x = 1, u = 2\).

\(V = \pi \int_{5}^{2} \frac{16}{u^2} \left(-\frac{1}{3}\right) \, du\).

\(V = -\frac{16\pi}{3} \int_{5}^{2} \frac{1}{u^2} \, du\).

\(V = -\frac{16\pi}{3} \left[ -\frac{1}{u} \right]_{5}^{2}\).

\(V = \frac{16\pi}{3} \left( \frac{1}{2} - \frac{1}{5} \right)\).

\(V = \frac{16\pi}{3} \times \frac{3}{10} = \frac{8\pi}{5}\).

(a) To find the equation of the line \(AB\), first calculate the gradient:

\(\text{Gradient of } AB = \frac{2 - (-1)}{5 - 2} = 1\)

Using point \(A(5, 2)\), the equation of the line is:

\(y - 2 = 1(x - 5)\)

Simplifying gives:

\(y = x - 3\)

(b) To find the volume of revolution, calculate the integral of the area between the curve \(x = y^2 + 1\) and the line \(y = x - 3\) rotated about the \(y\)-axis:

For the curve:

\(\pi \int (y^2 + 1)^2 \, dy = \pi \int (y^4 + 2y^2 + 1) \, dy\)

Integrating gives:

\(\pi \left( \frac{y^5}{5} + \frac{2y^3}{3} + y \right)\)

For the line:

\(\pi \int (y + 3)^2 \, dy = \pi \int (y^2 + 6y + 9) \, dy\)

Integrating gives:

\(\pi \left( \frac{y^3}{3} + 3y^2 + 9y \right)\)

Apply limits from \(y = -1\) to \(y = 2\):

\(\pi \left( \frac{8}{3} + 12 + 18 \right) - \pi \left( -\frac{1}{3} + 3 - 9 \right)\)

Calculating gives:

\(\pi \left( 39 - 15 \frac{3}{5} \right)\)

Volume = \(\frac{117}{5} \pi\) or approximately 73.5

(i) Differentiate \(y = (kx - 3)^{-1} + (kx - 3)\) to find \(\frac{dy}{dx} = -k(kx - 3)^{-2} + k\). Set \(\frac{dy}{dx} = 0\) to find stationary points:

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

\((kx - 3)^{2} = 1\)

\(k^{2}x^{2} - 6kx + 8k = 0\)

Solving gives \(x = \frac{2}{k}\) or \(x = \frac{4}{k}\).

To determine the nature, find \(\frac{d^{2}y}{dx^{2}} = 2k^{2}(kx - 3)^{-3}\).

When \(x = \frac{2}{k}\), \(\frac{d^{2}y}{dx^{2}} = (-2k^{2}) < 0\), so it is a maximum.

When \(x = \frac{4}{k}\), \(\frac{d^{2}y}{dx^{2}} = (2k^{2}) > 0\), so it is a minimum.

(ii) For \(k = 1\), the curve is \(y = (x - 3)^{-1} + (x - 3)\). The volume \(V\) is given by:

\(V = \pi \int_{0}^{2} [(x - 3)^{-1} + (x - 3)]^{2} \, dx\)

\(= \pi \int_{0}^{2} [(x - 3)^{-2} + (x - 3)^{2} + 2] \, dx\)

\(= \pi \left[ -(x - 3)^{-1} + \frac{(x - 3)^{3}}{3} + 2x \right]_{0}^{2}\)

\(= \pi \left[ -\frac{1}{3} + 4 - \left(-\frac{1}{3} - 9 + 0\right) \right]\)

\(= \frac{40\pi}{3}\) or 41.9

To find the volume of the solid of revolution, we use the formula:

\(V = \pi \int_{a}^{b} y^2 \, dx\)

Given \(y = (x^3 + 1)^{\frac{1}{2}}\), we have:

\(y^2 = x^3 + 1\)

Thus, the volume is:

\(V = \pi \int_{0}^{2} (x^3 + 1) \, dx\)

Integrating, we get:

\(\pi \left[ \frac{x^4}{4} + x \right]_{0}^{2}\)

Evaluating the definite integral:

\(\pi \left( \left( \frac{2^4}{4} + 2 \right) - \left( \frac{0^4}{4} + 0 \right) \right)\)

\(= \pi \left( 4 + 2 \right)\)

\(= 6\pi\)

Thus, the volume is \(6\pi\) or approximately 18.8.

(i) To find \(\frac{dy}{dx}\), differentiate \(y = \frac{8}{x} + 2x\):

\(\frac{dy}{dx} = -8x^{-2} + 2\).

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\):

\(\frac{d^2y}{dx^2} = 16x^{-3}\).

To find \(\int y^2 \, dx\), integrate \(y^2 = \left(\frac{8}{x} + 2x\right)^2\):

\(\int y^2 \, dx = -64x^{-1} + 32x + \frac{4x^3}{3} + C\).

(ii) Set \(\frac{dy}{dx} = 0\) to find stationary points:

\(-8x^{-2} + 2 = 0 \Rightarrow x = \pm 2\).

For \(x > 0\), \(M(2, 8)\).

For \(x < 0\), the other turning point is \((-2, -8)\).

Check the nature using \(\frac{d^2y}{dx^2}\):

At \(x = -2\), \(\frac{d^2y}{dx^2} < 0\), so it is a maximum.

(iii) The volume of revolution is given by:

\(V = \pi \int_{1}^{2} y^2 \, dx\).

Using the integral from part (i),

\(V = \pi \left[ -64x^{-1} + 32x + \frac{4x^3}{3} \right]_{1}^{2}\).

Calculate the definite integral:

\(V = \frac{220\pi}{3}\).