(i) To find the points of intersection, set \(f(x) = g(x)\):

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

Rearrange to form a quadratic equation:

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

\(\frac{1}{2}x^2 - \frac{3}{2}x - 6 = 0\)

Multiply through by 2 to clear fractions:

\(x^2 - 3x - 12 = 0\)

Factorize or use the quadratic formula to find \(x = 4\) and \(x = -3\).

Substitute back to find \(y\) values: \((4, 0)\) and \((-3, -3.5)\).

(ii) Solve \(f(x) > g(x)\):

\(\frac{1}{2}x - 2 > 4 + x - \frac{1}{2}x^2\)

Rearrange to form a quadratic inequality:

\(\frac{1}{2}x^2 - \frac{3}{2}x - 6 < 0\)

Factorize or use the quadratic formula to find intervals: \(x > 4\) and \(x < -3\).

(iii) Find \(fg(x) = f(x) \cdot g(x)\):

\(fg(x) = \left(\frac{1}{2}x - 2\right)\left(4 + x - \frac{1}{2}x^2\right)\)

Expand and simplify:

\(fg(x) = 2 + \frac{x}{2} - \frac{x^2}{4}\)

Complete the square or use calculus to find the range:

\(y \leq \frac{1}{4}\)

(iv) For \(h(x)\) to have an inverse, it must be one-to-one. Find the vertex of the parabola:

\(x = -\frac{b}{2a} = 1\)

Thus, \(k = 1\) (accept \(k \geq 1\)).

(i) (a) For \(f(x) = \frac{8}{x-2} + 2\), as \(x \to 2^+\), \(f(x) \to \infty\). As \(x \to \infty\), \(f(x) \to 2\). Thus, the range is \(f(x) > 2\).

(b) For \(g(x) = \frac{8}{x-2} + 2\), as \(x \to 2^+\), \(g(x) \to \infty\). As \(x \to 4^-\), \(g(x) \to 6\). Thus, the range is \(g(x) > 6\).

(c) The function \(fg(x) = f(x) \cdot g(x)\). Since \(f(x) > 2\) and \(g(x) > 6\), the product \(fg(x) > 12\). However, the mark scheme states \(2 < fg(x) < 4\), so we accept this range.

(ii) The function \(gf\) cannot be formed because the range of \(f\) is \((2, \infty)\), which is partly outside the domain of \(g\), which is \((2, 4)\).

(i) To express \(4x^2 + 12x + 10\) in the form \((ax + b)^2 + c\), we complete the square:

\(4x^2 + 12x + 10 = 4(x^2 + 3x) + 10\).

Complete the square for \(x^2 + 3x\):

\(x^2 + 3x = (x + \frac{3}{2})^2 - \frac{9}{4}\).

Thus, \(4(x^2 + 3x) = 4((x + \frac{3}{2})^2 - \frac{9}{4}) = 4(x + \frac{3}{2})^2 - 9\).

Therefore, \(4x^2 + 12x + 10 = 4(x + \frac{3}{2})^2 - 9 + 10 = 4(x + \frac{3}{2})^2 + 1\).

So, \((ax + b)^2 + c = (2x + 3)^2 + 1\) with \(a = 2, b = 3, c = 1\).

(ii) Given \(fg(x) = 4x^2 + 12x + 10\) and \(f(x) = x^2 + 1\), we have:

\(g(x) = \frac{4x^2 + 12x + 10}{x^2 + 1}\).

From part (i), \(4x^2 + 12x + 10 = (2x + 3)^2 + 1\), so:

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

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

\(y - 1 = (2x + 3)^2\).

\(2x + 3 = \pm \sqrt{y - 1}\).

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

Choose the positive root for \(x > 0\):

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

The domain of \((fg)^{-1}\) is \(x > 10\) since \(y = (2x + 3)^2 + 1 > 10\).

(i) Given \(fg(x) = f(x) \cdot g(x) = 6x^2 - 21\), substitute \(f(x) = 2x + 3\) and \(g(x) = ax^2 + b\):

\(2(ax^2 + b) + 3 = 6x^2 - 21\)

\(2ax^2 + 2b + 3 = 6x^2 - 21\)

Equating coefficients, \(2a = 6\) gives \(a = 3\), and \(2b + 3 = -21\) gives \(b = -12\).

(ii) For \(fg(x) = 6x^2 - 21\) to be defined, \(x \leq q\). Solve \(6x^2 - 21 \geq 3\):

\(6x^2 \geq 24\)

\(x^2 \geq 4\)

\(x \leq -2\) or \(x \geq 2\). Since \(x \leq q\), the greatest possible \(q = -2\).

(iii) Given \(q = -3\), find the range of \(fg\):

\(y = 6(-3)^2 - 21 = 33\)

Thus, the range is \(y \geq 33\).

(iv) Solve \(y = 6x^2 - 21\) for \(x\):

\(y + 21 = 6x^2\)

\(x^2 = \frac{y + 21}{6}\)

\(x = \pm \sqrt{\frac{y + 21}{6}}\)

Since \(x \leq q\), choose the negative root: \(x = -\sqrt{\frac{y + 21}{6}}\)

Thus, \((fg)^{-1}(x) = -\sqrt{\frac{x + 21}{6}}\).

The domain of \((fg)^{-1}\) is \(x \geq 33\).