First, find the inverse of \(f\). Given \(f(x) = 3x + 2\), set \(y = 3x + 2\).

Solve for \(x\):

\(y - 2 = 3x\)

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

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

Next, find \(gf(x)\):

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

Substitute into \(g\):

\(g(3x + 2) = 4(3x + 2) - 12\)

\(= 12x + 8 - 12\)

\(= 12x - 4\)

Equate \(f^{-1}(x)\) and \(gf(x)\):

\(\frac{x - 2}{3} = 12x - 4\)

Multiply through by 3 to clear the fraction:

\(x - 2 = 36x - 12\)

Rearrange and solve for \(x\):

\(x - 36x = -12 + 2\)

\(-35x = -10\)

\(x = \frac{2}{7}\)

(i) To solve \(ff(x) = 11\), first find \(f(f(x))\):

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

\(f(f(x)) = f(2x - 3) = 2(2x - 3) - 3 = 4x - 6 - 3 = 4x - 9\)

Set \(4x - 9 = 11\):

\(4x = 20\)

\(x = 5\)

(ii) The function \(g(x) = x^2 + 4x\) is a quadratic function. The vertex form is \(g(x) = (x + 2)^2 - 4\), with a minimum at \(x = -2\).

The minimum value is \(-4\), so the range is \(g(x) \geq -4\).

(iii) To find where \(g(x) > 12\):

\(x^2 + 4x > 12\)

\(x^2 + 4x - 12 > 0\)

Factor the quadratic: \((x - 2)(x + 6) > 0\)

The critical points are \(x = 2\) and \(x = -6\).

Test intervals: \(x < -6\), \(-6 < x < 2\), \(x > 2\).

The solution is \(x < -6\) or \(x > 2\).

(i) To find the inverse function \(f^{-1}\), start with \(y = x^2 + 1\).

Rearrange to solve for \(x\):

\(x^2 = y - 1\)

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

Since \(x \geq 0\), we take the positive root:

\(x = \sqrt{y - 1}\)

Interchange \(x\) and \(y\) to get the inverse function:

\(f^{-1} : x \mapsto \sqrt{x-1}\) for \(x > 1\).

(ii) Solve \(ff(x) = \frac{185}{16}\).

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

\(f(x) = x^2 + 1\)

Then, \(ff(x) = (x^2 + 1)^2 + 1\)

Set \((x^2 + 1)^2 + 1 = \frac{185}{16}\)

\((x^2 + 1)^2 = \frac{185}{16} - 1 = \frac{169}{16}\)

\(x^2 + 1 = \pm \frac{13}{4}\)

Since \(x^2 + 1 \geq 1\), take the positive root:

\(x^2 + 1 = \frac{13}{4}\)

\(x^2 = \frac{13}{4} - 1 = \frac{9}{4}\)

\(x = \pm \frac{3}{2}\)

Since \(x \geq 0\), \(x = \frac{3}{2}\).

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

Solve for \(x\):

\(y - 3 = 2x\)

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

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

To solve \(f(x) = f^{-1}(x)\):

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

Multiply through by 2 to clear the fraction:

\(4x + 6 = x - 3\)

\(3x = -9\)

\(x = -3\)

(iii) Find \(gf(x) \leq 16\):

\(gf(x) = g(2x + 3) = (2x + 3)^2 - 6(2x + 3)\)

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

\(= 4x^2 - 9\)

Set \(4x^2 - 9 \leq 16\):

\(4x^2 \leq 25\)

\(x^2 \leq \frac{25}{4}\)

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

Given \(x \leq 0\), the solution is \(-\frac{5}{2} \leq x \leq 0\).

(i) To find \(a\), we use \(ff(2) = 10\). First, calculate \(f(2) = 3(2) + a = 6 + a\). Then, \(ff(2) = f(f(2)) = f(6 + a) = 3(6 + a) + a = 18 + 3a + a = 18 + 4a\). Given \(ff(2) = 10\), we have:

\(18 + 4a = 10\)

\(4a = 10 - 18\)

\(4a = -8\)

\(a = -2\)

To find \(b\), use \(g^{-1}(2) = 3\). The inverse function \(g^{-1}(x)\) is found by solving \(y = b - 2x\) for \(x\):

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

Given \(g^{-1}(2) = 3\), we have:

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

\(6 = b - 2\)

\(b = 8\)

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

\(fg(x) = f(g(x)) = f(b - 2x) = 3(b - 2x) + a\)

Substitute \(a = -2\) and \(b = 8\):

\(fg(x) = 3(8 - 2x) - 2\)

\(fg(x) = 24 - 6x - 2\)

\(fg(x) = 22 - 6x\)