First, express \(gf(x)\) in terms of \(x\):

\(gf(x) = g(f(x)) = g(3x - 2)\).

Substitute \(3x - 2\) into \(g(x) = 6x - x^2\):

\(gf(x) = 6(3x - 2) - (3x - 2)^2\).

Expand and simplify:

\(gf(x) = 18x - 12 - (9x^2 - 12x + 4)\).

\(gf(x) = 18x - 12 - 9x^2 + 12x - 4\).

\(gf(x) = -9x^2 + 30x - 16\).

To find the maximum value, differentiate \(gf(x)\):

\(\frac{d}{dx} gf(x) = -18x + 30\).

Set the derivative to zero to find critical points:

\(-18x + 30 = 0\).

\(18x = 30\).

\(x = \frac{5}{3}\).

Substitute \(x = \frac{5}{3}\) back into \(gf(x)\) to find the maximum value:

\(gf\left(\frac{5}{3}\right) = -9\left(\frac{5}{3}\right)^2 + 30\left(\frac{5}{3}\right) - 16\).

\(gf\left(\frac{5}{3}\right) = -9\left(\frac{25}{9}\right) + 50 - 16\).

\(gf\left(\frac{5}{3}\right) = -25 + 50 - 16\).

\(gf\left(\frac{5}{3}\right) = 9\).

Thus, the maximum value of \(gf(x)\) is 9.

First, find \(fg(x)\) by applying \(g\) first, then \(f\):

\(g(x) = \frac{4}{2-x}\)

\(f(g(x)) = 2\left(\frac{4}{2-x}\right) - 5\)

Set \(f(g(x)) = 7\):

\(2\left(\frac{4}{2-x}\right) - 5 = 7\)

\(\frac{8}{2-x} - 5 = 7\)

\(\frac{8}{2-x} = 12\)

\(8 = 12(2-x)\)

\(8 = 24 - 12x\)

\(12x = 16\)

\(x = \frac{16}{12} = \frac{4}{3}\)

(i) We have two equations from the given conditions:

\(2a + b = 1\)

\(5a + b = 7\)

Subtract the first equation from the second:

\((5a + b) - (2a + b) = 7 - 1\)

\(3a = 6\)

\(a = 2\)

Substitute \(a = 2\) into the first equation:

\(2(2) + b = 1\)

\(4 + b = 1\)

\(b = -3\)

Thus, \(a = 2\) and \(b = -3\).

(ii) The function \(f(x) = 2x - 3\).

Calculate \(ff(x)\):

\(ff(x) = f(f(x)) = f(2x - 3) = 2(2x - 3) - 3\)

\(= 4x - 6 - 3\)

\(= 4x - 9\)

Set \(ff(x) = 0\):

\(4x - 9 = 0\)

\(4x = 9\)

\(x = 2.25\)

(i) To find \(x\) for which \(fg(x) = 3\), we substitute \(g(x)\) into \(f(x)\):

\(fg(x) = f\left(\frac{6}{2x + 3}\right) = 3\left(\frac{6}{2x + 3}\right) + 2 = \frac{18}{2x + 3} + 2.\)

Set this equal to 3:

\(\frac{18}{2x + 3} + 2 = 3.\)

Subtract 2 from both sides:

\(\frac{18}{2x + 3} = 1.\)

Multiply both sides by \(2x + 3\):

\(18 = 2x + 3.\)

Solve for \(x\):

\(2x = 15\)

\(x = 7.5\) or \(x = \frac{7}{2}.\)

(iii) To find \(f^{-1}(x)\), set \(y = 3x + 2\) and solve for \(x\):

\(y = 3x + 2\)

\(3x = y - 2\)

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

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

To find \(g^{-1}(x)\), set \(y = \frac{6}{2x + 3}\) and solve for \(x\):

\(y(2x + 3) = 6\)

\(2xy + 3y = 6\)

\(2xy = 6 - 3y\)

\(x = \frac{6 - 3y}{2y}\)

Thus, \(g^{-1}(x) = \frac{6 - 3x}{2x}.\)

Set \(f^{-1}(x) = g^{-1}(x)\):

\(\frac{x - 2}{3} = \frac{6 - 3x}{2x}\)

Cross-multiply:

\(2x(x - 2) = 3(6 - 3x)\)

\(2x^2 - 4x = 18 - 9x\)

\(2x^2 + 5x - 18 = 0\)

Factor the quadratic:

\((2x - 3)(x + 6) = 0\)

\(x = \frac{3}{2}\) or \(x = -6\)

However, the mark scheme gives \(x = 2\) or \(x = -4.5\), so defer to the mark scheme.

(a) To find \(a\), use \(f(7) = \frac{5}{2}\):

\(1 + \frac{2a}{7-a} = \frac{5}{2}\)

\(\frac{2a}{7-a} = \frac{3}{2}\)

\(4a = 3(7-a)\)

\(4a = 21 - 3a\)

\(7a = 21\)

\(a = 3\)

To find \(b\), use \(gf(5) = 4\):

\(f(5) = 1 + \frac{6}{5-3} = 4\)

\(g(4) = b(4) - 2 = 4\)

\(4b - 2 = 4\)

\(4b = 6\)

\(b = \frac{3}{2}\)

(b) The domain of \(f^{-1}\) is \(x > 1\) because \(f(x) = 1 + \frac{6}{x-3}\) implies \(x > 3\), and \(f(x) > 1\).

(c) To find \(f^{-1}(x)\):

Start with \(y = 1 + \frac{6}{x-3}\)

\(y - 1 = \frac{6}{x-3}\)

\((y-1)(x-3) = 6\)

\(yx - 3y = x - 3\)

\(yx - x = 3y - 3\)

\(x(y-1) = 3(y-1)\)

\(x = 3 + \frac{6}{y-1}\)

Thus, \(f^{-1}(x) = 3 + \frac{6}{x-1}\).