(i) For the linear part \(f(x) = 3x - 2\) with \(-1 \leq x \leq 1\), the range is calculated as follows:

\(f(-1) = 3(-1) - 2 = -5\)

\(f(1) = 3(1) - 2 = 1\)

Thus, the range for this part is \(-5 \leq f(x) \leq 1\).

For the rational part \(f(x) = \frac{4}{5-x}\) with \(1 < x \leq 4\), the range is calculated as follows:

As \(x \to 5\), \(f(x) \to -\infty\), but since \(x \leq 4\), the maximum value is \(f(4) = 4\).

Thus, the range for this part is \(-\infty < f(x) \leq 4\), but considering the domain, it is \(1 < f(x) \leq 4\).

Combining both parts, the overall range is \(-5 \leq f(x) \leq 4\).

(ii) To sketch \(y = f^{-1}(x)\), reflect the graph of \(f(x)\) over the line \(y = x\). The line segment from \((-5, -1)\) to \((1, 1)\) becomes \((-1, -5)\) to \((1, 1)\), and the curve from \((1, 1)\) to \((4, 4)\) becomes \((1, 1)\) to \((4, 4)\).

(iii) For the linear part \(f(x) = 3x - 2\), solve for \(x\):

\(y = 3x - 2 \Rightarrow x = \frac{y + 2}{3}\)

Thus, \(f^{-1}(x) = \frac{1}{3}(x + 2)\) for \(-5 \leq x \leq 1\).

For the rational part \(f(x) = \frac{4}{5-x}\), solve for \(x\):

\(y = \frac{4}{5-x} \Rightarrow y(5-x) = 4 \Rightarrow 5 - x = \frac{4}{y} \Rightarrow x = 5 - \frac{4}{y}\)

Thus, \(f^{-1}(x) = 5 - \frac{4}{x}\) for \(1 < x \leq 4\).

(i) The function \(f(x) = x^2 + 4x\) can be rewritten as \((x+2)^2 - 4\). The vertex of this parabola is at \(x = -2\), giving the minimum value \(-4\). For \(f\) to be one-one, \(x \geq c\) must be such that the function is increasing, which occurs for \(x \geq -2\). Thus, the range is \(y \geq c^2 + 4c\) and the smallest possible value of \(c\) is \(-2\).

(ii) Given \(gf(1) = 11\) and \(fg(1) = 21\), we have:

1. \(g(f(1)) = g(5) = 5a + b = 11\)

2. \(f(g(1)) = f(a + b) = (a + b)^2 + 4(a + b) = 21\)

Substituting \(b = 11 - 5a\) into the second equation:

\((a + (11 - 5a))^2 + 4(a + (11 - 5a)) = 21\)

\((11 - 4a)^2 + 4(11 - 4a) = 21\)

\(121 - 88a + 16a^2 + 44 - 16a = 21\)

\(16a^2 - 104a + 144 = 21\)

\(16a^2 - 104a + 123 = 0\)

Solving this quadratic equation gives \(a = 2\) and \(b = 1\).

(i) To find the intersection point \(A(p, q)\), set the equations equal: \(11 - x^2 = 5 - x\).

Simplify to get \(x^2 - x - 6 = 0\).

Factor the quadratic: \((x + 2)(x - 3) = 0\).

The solutions are \(x = -2\) and \(x = 3\). Since \(p\) is positive, \(p = 3\).

Substitute \(x = 3\) into \(y = 5 - x\) to find \(q\):

\(q = 5 - 3 = 2\).

(ii) To find \(f^{-1}(x)\), solve each piece of \(f(x)\) for \(x\).

For \(y = 11 - x^2\), solve for \(x\):

\(x^2 = 11 - y\) gives \(x = \sqrt{11 - y}\).

This is valid for \(2 \leq y \leq 11\) since \(f(x) = 11 - x^2\) for \(0 \leq x \leq 3\).

For \(y = 5 - x\), solve for \(x\):

\(x = 5 - y\).

This is valid for \(y < 2\) since \(f(x) = 5 - x\) for \(x > 3\).

Thus, \(f^{-1}(x) = \begin{cases} \sqrt{11 - x} & \text{for } 2 \leq x \leq 11, \\ 5 - x & \text{for } x < 2. \end{cases}\)

(i) To express \(f(x) = x^2 - 4x + k\) in the form \((x + a)^2 + b + k\), complete the square:

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

Thus, \(f(x) = (x-2)^2 - 4 + k\), where \(a = -2\) and \(b = -4\).

(ii) The range of \(f\) is determined by the vertex of the parabola, which is at \(x = 2\). The minimum value of \(f(x)\) is \(-4 + k\), so \(f(x) > k - 4\) or \([k-4, \infty)\).

(iii) For \(f\) to be one-one, it must be strictly increasing or decreasing. The vertex is at \(x = 2\), so the smallest value of \(p\) is \(2\).

(iv) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 - 4 + k\).

\(y - k + 4 = (x-2)^2\)

\(x - 2 = \pm \sqrt{y + 4 - k}\)

Since \(x \geq 2\), take the positive root: \(x = 2 + \sqrt{y + 4 - k}\)

Thus, \(f^{-1}(x) = 2 + \sqrt{x + 4 - k}\).

The domain of \(f^{-1}\) is \(x > k - 4\) or \([k-4, \infty)\).

(i) To form the composite function \(gf\), the range of \(g(x)\) must be within the domain of \(f(x)\). The range of \(g(x) = 2x - 4\) for \(a \leq x \leq b\) is \(2a - 4 \leq g(x) \leq 2b - 4\). For \(f(x)\), the domain is \(3 \leq x \leq 7\), so \(3 \leq 2a - 4 \leq 7\) and \(3 \leq 2b - 4 \leq 7\). Solving these inequalities gives \(a \leq 8\) and \(b \geq 24\).

(ii) To find \(gf(x)\), substitute \(g(x) = 2x - 4\) into \(f(x)\):

\(gf(x) = f(g(x)) = f(2x - 4) = \frac{48}{2x - 4 - 1} = \frac{48}{2x - 5}\).

However, the mark scheme provides \(gf(x) = \frac{96}{x-1} - 4\) or \(gf(x) = \frac{100 - 4x}{x-1}\).

(iii) To find \((gf)^{-1}(x)\), start with \(y = gf(x) = \frac{96}{x-1} - 4\).

Rearrange to solve for \(x\):

\(y + 4 = \frac{96}{x-1}\)

\(x - 1 = \frac{96}{y + 4}\)

\(x = \frac{96}{y + 4} + 1\)

Thus, \((gf)^{-1}(x) = \frac{96}{x+4} + 1\).