(i) To find \(gg(x)\), substitute \(g(x) = 2x - 3\) into itself:

\(gg(x) = g(g(x)) = g(2x - 3) = 2(2x - 3) - 3 = 4x - 6 - 3 = 4x - 9\).

(ii) To find \(f^{-1}(x)\), start with \(y = \frac{1}{x^2 - 9}\).

Rearrange to express \(x\) in terms of \(y\):

\(y = \frac{1}{x^2 - 9} \Rightarrow x^2 - 9 = \frac{1}{y} \Rightarrow x^2 = \frac{1}{y} + 9\).

Taking the square root, \(x = \sqrt{\frac{1}{y} + 9}\).

Thus, \(f^{-1}(x) = \sqrt{\frac{1}{x} + 9}\).

The domain of \(f^{-1}\) is \(x > 0\) because \(y > 0\) in the original function.

(iii) Solve \(fg(x) = \frac{1}{7}\):

\(fg(x) = \frac{1}{(2x - 3)^2 - 9} = \frac{1}{7}\).

\((2x - 3)^2 - 9 = 7 \Rightarrow (2x - 3)^2 = 16\).

\(2x - 3 = \pm 4 \Rightarrow 2x = 7 \text{ or } 2x = -1 \Rightarrow x = \frac{7}{2} \text{ or } x = -\frac{1}{2}\).

Since \(x > 3\), the solution is \(x = \frac{7}{2}\).

(i) To find \(gf(x)\), substitute \(g(x)\) into \(f\):

\(gf(x) = f(g(x)) = f(3x + 2) = 2(3x + 2)^2 + 3\).

Expanding, \((3x + 2)^2 = 9x^2 + 12x + 4\), so:

\(gf(x) = 2(9x^2 + 12x + 4) + 3 = 18x^2 + 24x + 8 + 3 = 18x^2 + 24x + 11\).

Thus, \(gf(x) = 6x^2 + 11\).

For \(fg(x)\), substitute \(f(x)\) into \(g\):

\(fg(x) = g(f(x)) = g(2x^2 + 3) = 3(2x^2 + 3) + 2 = 6x^2 + 9 + 2 = 6x^2 + 11\).

(ii) To find \((fg)^{-1}(x)\), start with \(y = fg(x) = 2(3x + 2)^2 + 3\).

Subtract 3: \(y - 3 = 2(3x + 2)^2\).

Divide by 2: \(\frac{y - 3}{2} = (3x + 2)^2\).

Take the square root: \(3x + 2 = \pm \sqrt{\frac{y - 3}{2}}\).

Subtract 2: \(3x = \pm \sqrt{\frac{y - 3}{2}} - 2\).

Divide by 3: \(x = \frac{1}{3}(\pm \sqrt{\frac{y - 3}{2}} - 2)\).

Thus, \((fg)^{-1}(x) = \frac{1}{3}\sqrt{\frac{x-3}{2}} - \frac{2}{3}\).

The domain is \(x \geq 11\) since \(fg(x)\) is defined for \(x > 0\).

(iii) Solve \(gf(2x) = fg(x)\):

\(gf(2x) = 6(2x)^2 + 11 = 24x^2 + 11\).

\(fg(x) = 6x^2 + 11\).

Equate: \(24x^2 + 11 = 6x^2 + 11\).

Simplify: \(18x^2 = 0\).

Thus, \(x = 0\).

Check for other solutions: \(6x^2 - 24x = 0\).

Factor: \(6x(x - 4) = 0\).

Solutions: \(x = 0, 4\).

(b) To find the least possible value of \(p\) and the greatest possible value of \(q\), we need to ensure that the range of \(g(x)\) fits within the domain of \(f(x)\). The function \(g(x) = 3x + 1\) is defined for \(x < 8\), so \(g(x) < 25\). We need \(p < g(x) < q\).

Solving \(4x^2 - 12x + 13 < 8\), we get:

\((2x - 3)^2 + 4 < 8\)

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

\(-2 < 2x - 3 < 2\)

\(\frac{1}{2} < x < 2\frac{1}{2}\)

Thus, the least \(p = \frac{1}{2}\) and the greatest \(q = 2\frac{1}{2}\).

(c) To find \(gf(x)\), substitute \(g(x) = 3x + 1\) into \(f(x)\):

\(gf(x) = f(g(x)) = f(3x + 1)\)

\(gf(x) = 4(3x + 1)^2 - 12(3x + 1) + 13\)

\(gf(x) = 4(9x^2 + 6x + 1) - 36x - 12 + 13\)

\(gf(x) = 36x^2 + 24x + 4 - 36x - 12 + 13\)

\(gf(x) = 12x^2 - 36x + 40\)

(i) To find \(fg(x)\), multiply \(f(x)\) and \(g(x)\):

\(fg(x) = \left( \frac{4}{x} - 2 \right) \cdot \frac{4}{5x + 2}\).

Simplify:

\(fg(x) = \frac{4}{x} \cdot \frac{4}{5x + 2} - 2 \cdot \frac{4}{5x + 2}\).

\(fg(x) = \frac{16}{x(5x + 2)} - \frac{8}{5x + 2}\).

Combine the fractions:

\(fg(x) = \frac{16 - 8x}{x(5x + 2)}\).

Further simplification gives \(fg(x) = 5x\).

The range of \(fg\) is \(y \geq 0\).

(ii) To find \(g^{-1}(x)\), start with \(y = \frac{4}{5x + 2}\).

Solve for \(x\):

\(y(5x + 2) = 4\).

\(5xy + 2y = 4\).

\(5xy = 4 - 2y\).

\(x = \frac{4 - 2y}{5y}\).

Thus, \(g^{-1}(x) = \frac{4 - 2x}{5x}\).

The domain of \(g^{-1}\) is \(0 < x \leq 2\).

First, calculate \(ff(x)\):

\(ff(x) = f(f(x)) = f(10 - 3x) = 10 - 3(10 - 3x)\)

\(= 10 - 30 + 9x = 9x - 20\).

Next, calculate \(gf(2)\):

\(f(2) = 10 - 3(2) = 10 - 6 = 4\).

\(gf(2) = g(f(2)) = g(4) = \frac{10}{3 - 2(4)} = \frac{10}{3 - 8} = \frac{10}{-5} = -2\).

Set \(ff(x) = gf(2)\):

\(9x - 20 = -2\).

Solve for \(x\):

\(9x = 18\)

\(x = 2\).