(i) To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = 2(3x + 4)^{\frac{3}{2}} - 6x - 8\) with respect to \(x\).

The derivative of \(2(3x + 4)^{\frac{3}{2}}\) is \(2 \times \frac{3}{2} \times (3x + 4)^{\frac{1}{2}} \times 3 = 9(3x + 4)^{\frac{1}{2}}\).

The derivative of \(-6x\) is \(-6\), and the derivative of \(-8\) is \(0\).

Thus, \(\frac{d^2y}{dx^2} = 9(3x + 4)^{\frac{1}{2}} - 6\).

(ii) To verify the stationary point at \(x = -1\), substitute \(x = -1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2(3(-1) + 4)^{\frac{3}{2}} - 6(-1) - 8 = 2(1)^{\frac{3}{2}} + 6 - 8 = 0\).

Since \(\frac{dy}{dx} = 0\), there is a stationary point at \(x = -1\).

To determine the nature, evaluate \(\frac{d^2y}{dx^2}\) at \(x = -1\):

\(\frac{d^2y}{dx^2} = 9(3(-1) + 4)^{\frac{1}{2}} - 6 = 9(1)^{\frac{1}{2}} - 6 = 3\).

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.

To find the stationary point, we first differentiate the equation \(y = 2x + \frac{1}{(x-1)^2}\).

The derivative is \(\frac{dy}{dx} = 2 - 2(x-1)^{-3}\).

Substitute \(x = 2\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2 - 2(2-1)^{-3} = 2 - 2 = 0\).

This confirms a stationary point at \(x = 2\).

Next, we find the second derivative to determine the nature of the stationary point:

\(\frac{d^2y}{dx^2} = 6(x-1)^{-4}\).

Substitute \(x = 2\) into \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = 6(2-1)^{-4} = 6\).

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point at \(x = 2\) is a minimum.

(i) To find where the gradient is less than 5, first find the derivative: \(f'(x) = 3x^2 - 4x + 1\).

Set \(f'(x) < 5\):

\(3x^2 - 4x + 1 < 5\)

\(3x^2 - 4x - 4 < 0\)

Factorize: \((3x + 2)(x - 2) < 0\)

Find the critical points: \(x = -\frac{2}{3}\) and \(x = 2\).

The solution is \(-\frac{2}{3} < x < 2\).

(ii) To find stationary points, set \(f'(x) = 0\):

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

Factorize: \((3x - 1)(x - 1) = 0\)

Solutions: \(x = \frac{1}{3}\) and \(x = 1\).

Find \(f(x)\) at these points:

\(f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - 2\left(\frac{1}{3}\right)^2 + \frac{1}{3} = \frac{4}{27}\)

\(f(1) = 1^3 - 2 \times 1^2 + 1 = 0\)

Determine nature using second derivative: \(f''(x) = 6x - 4\).

\(f''\left(\frac{1}{3}\right) = 6\left(\frac{1}{3}\right) - 4 = -2\) (negative, so maximum)

\(f''(1) = 6 \times 1 - 4 = 2\) (positive, so minimum)

Maximum at \(\left( \frac{1}{3}, \frac{4}{27} \right)\); Minimum at \((1, 0)\).

To find the gradient of the curve, we need to differentiate the equation \(y = 3x^3 - 6x^2 + 4x + 2\) with respect to \(x\).

The derivative is \(\frac{dy}{dx} = 9x^2 - 12x + 4\).

We need to show that \(9x^2 - 12x + 4 \geq 0\) for all \(x\).

Consider the expression \(9x^2 - 12x + 4\).

This can be rewritten as \((3x - 2)^2\).

Since \((3x - 2)^2\) is a square, it is always non-negative.

Therefore, \(9x^2 - 12x + 4 \geq 0\) for all \(x\), and the gradient of the curve is never negative.

To find \(g'(x)\), differentiate \(g(x) = 2(x-1)^3 + 8\).

Using the chain rule, \(\frac{d}{dx}[(x-1)^3] = 3(x-1)^2 \cdot 1\).

Thus, \(g'(x) = 2 \cdot 3(x-1)^2 = 6(x-1)^2\).

Since \(g'(x) = 6(x-1)^2 > 0\) for \(x > 1\), \(g(x)\) is strictly increasing.

This implies \(g\) is one-to-one (1:1), so \(g\) has an inverse.