(i) To find the largest value of \(a\) for which the composite function \(gf\) can be formed, we need \(3x + 1 \leq -1\). Solving for \(x\), we have:

\(3x + 1 \leq -1\)

\(3x \leq -2\)

\(x \leq -\frac{2}{3}\)

Thus, the largest value of \(a\) is \(-\frac{2}{3}\).

(ii) For \(a = -1\), solve \(fg(x) + 14 = 0\):

\(fg(x) = 3(-1 - x^2) + 1\)

\(fg(x) + 14 = 0 \Rightarrow 3(-1 - x^2) + 1 + 14 = 0\)

\(3(-1 - x^2) + 15 = 0 \Rightarrow 3(-1 - x^2) = -15\)

\(-3 - 3x^2 = -15 \Rightarrow 3x^2 = 12\)

\(x^2 = 4 \Rightarrow x = -2\) (since \(x \leq -1\))

(iii) Find the set of values of \(x\) which satisfy \(gf(x) \leq -50\):

\(gf(x) = -1 - (3x + 1)^2\)

\(gf(x) \leq -50 \Rightarrow -1 - (3x + 1)^2 \leq -50\)

\(-(3x + 1)^2 \leq -49 \Rightarrow (3x + 1)^2 \geq 49\)

\(3x + 1 \geq 7\) or \(3x + 1 \leq -7\)

\(3x \geq 6 \Rightarrow x \geq 2\) or \(3x \leq -8 \Rightarrow x \leq -\frac{8}{3}\)

Since \(x \leq -1\), the solution is \(x \leq -\frac{8}{3}\).

(i) Completing the square for \(f(x) = x^2 + 6x - 8\), we have:

\(x^2 + 6x - 8 = (x+3)^2 - 17\)

The minimum value of \((x+3)^2\) is 0, so the minimum value of \(f(x)\) is \(-17\). Thus, the range of \(f(x)\) is \(\geq -17\).

(ii) Given roots \(x = k\) and \(x = -2k\), the equation can be written as:

\((x-k)(x+2k) = 0\)

Expanding gives \(x^2 + 5kx - 2k^2 = 0\). Comparing with \(x^2 + 5x + b = 0\), we find:

\(5k = 5 \Rightarrow k = 5\)

\(b = -2k^2 = -2(5)^2 = -50\)

(iii) For \(f(x+a) = a\), substitute \(x+a\) into \(f\):

\((x+a)^2 + a(x+a) + b = a\)

\(x^2 + 2ax + a^2 + ax + a^2 + b = a\)

\(x^2 + (2a+a)x + (a^2 + b - a) = 0\)

Using the discriminant \(b^2 - 4ac\) for no real roots:

\((3a)^2 - 4(1)(a^2 + b - a) < 0\)

\(9a^2 - 4(a^2 + b - a) < 0\)

\(9a^2 - 4a^2 - 4b + 4a < 0\)

\(5a^2 < 4(b-a)\)

\(a^2 < 4(b-a)\)

(i) To find \(f(x)\), we start with the inverse function \(f^{-1}(x) = \frac{1 - 5x}{2x}\). We need to solve for \(x\) in terms of \(y\):

\(y = \frac{1 - 5x}{2x}\)

\(2xy = 1 - 5x\)

\(2xy + 5x = 1\)

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

\(x = \frac{1}{2y + 5}\)

Thus, \(f(x) = \frac{1}{2x + 5}\).

The domain of \(f\) is determined by the range of \(f^{-1}\), which is \(x > -\frac{9}{4}\).

(ii) To find \(f^{-1}g(x)\), substitute \(g(x) = \frac{1}{x}\) into \(f^{-1}\):

\(f^{-1}\left(\frac{1}{x}\right) = \frac{1 - 5\left(\frac{1}{x}\right)}{2\left(\frac{1}{x}\right)}\)

\(= \frac{x - 5}{2}\)

Thus, \(f^{-1}g(x) = \frac{x - 5}{2}\).

(i) Given \(fg(1) = 2\), we have \(f(g(1)) = f(1^2) = f(1) = (a \cdot 1 + b)^{\frac{1}{3}} = 2\).

This implies \((a + b)^{\frac{1}{3}} = 2\), so \(a + b = 8\).

Given \(gf(9) = 16\), we have \(g(f(9)) = g((9a + b)^{\frac{1}{3}}) = ((9a + b)^{\frac{1}{3}})^2 = 16\).

This implies \((9a + b)^{\frac{2}{3}} = 16\), so \(9a + b = 64\).

Solving the equations \(a + b = 8\) and \(9a + b = 64\), we find \(a = 7\) and \(b = 1\).

(ii) To find \(f^{-1}(x)\), start with \(x = (7y + 1)^{\frac{1}{3}}\).

Cubing both sides gives \(x^3 = 7y + 1\).

Solving for \(y\), we get \(y = \frac{1}{7}(x^3 - 1)\).

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

The domain of \(f^{-1}\) is \(x \geq 1\) since \(f(x)\) is defined for \(x \geq 0\) and \(f(0) = 1\).

(i) To express \(x^2 - 2x - 15\) in the form \((x + a)^2 + b\), complete the square:

\(x^2 - 2x - 15 = (x-1)^2 - 1 - 15 = (x-1)^2 - 16\).

(ii) The smallest possible value of \(c\) is the minimum value of \((x-1)^2 - 16\), which is \(-16\).

(iii) Given \(c = 9\) and \(d = 65\), solve:

\(9 \leq (x-1)^2 - 16 \leq 65\)

\(25 \leq (x-1)^2 \leq 81\)

\(5 \leq x-1 \leq 9\)

\(6 \leq x \leq 10\)

Thus, \(p = 6\) and \(q = 10\).

(iv) To find \(f^{-1}(x)\), start with \(x = (y-1)^2 - 16\):

\(x + 16 = (y-1)^2\)

\(y - 1 = \pm \sqrt{x + 16}\)

Since \(y \geq 1\), \(y - 1 = \sqrt{x + 16}\)

\(y = 1 + \sqrt{x + 16}\)

Thus, \(f^{-1}(x) = 1 + \sqrt{x + 16}\).