To find \(f'(x)\), we differentiate \(f(x) = 2 - \frac{3}{4x-p}\).

Using the chain rule, let \(u = 4x - p\), then \(\frac{du}{dx} = 4\).

The derivative of \(\frac{3}{u}\) with respect to \(u\) is \(-\frac{3}{u^2}\).

Thus, \(\frac{d}{dx} \left( \frac{3}{4x-p} \right) = -3 \cdot \frac{1}{(4x-p)^2} \cdot 4 = \frac{-12}{(4x-p)^2}\).

Therefore, \(f'(x) = 0 - \left( \frac{-12}{(4x-p)^2} \right) = \frac{12}{(4x-p)^2}\).

Since \(\frac{12}{(4x-p)^2} > 0\) for \(x > \frac{p}{4}\), \(f\) is an increasing function.

To determine when the function \(f(x) = x^3 - x^2 - 8x + 5\) is increasing, we need to find where its derivative is positive.

First, differentiate \(f(x)\):

\(f'(x) = 3x^2 - 2x - 8\).

Set \(f'(x) > 0\) to find where the function is increasing:

\(3x^2 - 2x - 8 > 0\).

Solving the inequality \(3x^2 - 2x - 8 = 0\) gives the critical points. Using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -2\), \(c = -8\).

\(x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 3 \times (-8)}}{2 \times 3}\)

\(x = \frac{2 \pm \sqrt{4 + 96}}{6}\)

\(x = \frac{2 \pm \sqrt{100}}{6}\)

\(x = \frac{2 \pm 10}{6}\)

\(x = 2\) or \(x = -\frac{4}{3}\).

The function is increasing in the interval \(x < -\frac{4}{3}\) and \(x > 2\).

Since \(x < a\), the largest possible value of \(a\) is \(-\frac{4}{3}\).

To determine when the function \(f(x) = x^3 - 3x^2 - 9x + 2\) is increasing, we need to find when its derivative \(f'(x)\) is non-negative.

First, find the derivative:

\(f'(x) = 3x^2 - 6x - 9\).

Set \(f'(x) \geq 0\) to find the critical points:

\(3x^2 - 6x - 9 = 0\).

Factor the quadratic equation:

\(3(x^2 - 2x - 3) = 0\).

\(3(x - 3)(x + 1) = 0\).

The critical points are \(x = 3\) and \(x = -1\).

Since \(f(x)\) is increasing for \(x > n\), we need \(f'(x) \geq 0\) for \(x > n\).

Test intervals around the critical points:

- For \(x > 3\), \(f'(x) > 0\).

- For \(-1 < x < 3\), \(f'(x) < 0\).

- For \(x < -1\), \(f'(x) > 0\).

Thus, \(f(x)\) is increasing for \(x > 3\).

Therefore, the least possible value of \(n\) is 3.

(i) To express \(3x^2 - 6x + 2\) in the form \(a(x+b)^2 + c\), we complete the square:

Start with \(3(x^2 - 2x) + 2\).

Complete the square for \(x^2 - 2x\):

\(x^2 - 2x = (x-1)^2 - 1\).

Substitute back: \(3((x-1)^2 - 1) + 2 = 3(x-1)^2 - 3 + 2 = 3(x-1)^2 - 1\).

Thus, \(a = 3\), \(b = -1\), \(c = -1\).

(ii) Differentiate \(f(x) = x^3 - 3x^2 + 7x - 8\):

\(f'(x) = 3x^2 - 6x + 7\).

Complete the square for \(3x^2 - 6x + 7\):

\(3(x^2 - 2x) + 7 = 3((x-1)^2 - 1) + 7 = 3(x-1)^2 + 4\).

Since \(3(x-1)^2 + 4 > 0\) for all \(x\), \(f'(x) > 0\) for all \(x\).

Therefore, \(f\) is an increasing function.

(i) To find \(f'(x)\), differentiate \(f(x) = \frac{1}{x+1} + \frac{1}{(x+1)^2}\).

The derivative of \(\frac{1}{x+1}\) is \(-(x+1)^{-2}\).

The derivative of \(\frac{1}{(x+1)^2}\) is \(-2(x+1)^{-3}\).

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

(ii) Since \(f'(x) < 0\) for \(x > -1\), the function \(f\) is decreasing.

(iii) To find the stationary point of \(g(x) = \frac{1}{x+1} + \frac{1}{(x+1)^2}\), set \(\frac{dy}{dx} = 0\).

\(-\frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} = 0\) or \(-\frac{x^2 - 4x - 3}{(x+1)^4} = 0\).

Multiply by \((x+1)^3\) to get \(-(x+1) - 2 = 0\) or \(-x^2 - 4x - 3 = 0\).

Solving \(-x^2 - 4x - 3 = 0\) gives \(x = -3\).

Substitute \(x = -3\) into \(g(x)\) to find \(y = -1/4\).

Thus, the coordinates of the stationary point are \((-3, -1/4)\).