(i) For \(0 \leq x \leq 2\), \(f(x) = \frac{1}{2}x^2\). The minimum value is \(0\) and the maximum value is \(2\). Thus, the range is \(0 \leq f(x) \leq 2\).

For \(2 < x \leq 6\), \(f(x) = \frac{1}{2}x + 1\). The minimum value is \(2\) and the maximum value is \(4\). Thus, the range is \(2 < f(x) \leq 4\).

Combining both parts, the range of \(f\) is \(0 < f(x) < 4\).

(ii) To sketch \(y = f^{-1}(x)\), reflect the graph of \(f\) across the line \(y = x\).

(iii) For \(0 < x < 2\), solve \(y = \frac{1}{2}x^2\) for \(x\):

\(x = \sqrt{2y}\).

For \(2 < x < 4\), solve \(y = \frac{1}{2}x + 1\) for \(x\):

\(x = 2y - 2\).

Thus, \(f^{-1}(x)\) is defined by \(x \mapsto \sqrt{2x}\) for \(0 < x < 2\) and \(x \mapsto 2x - 2\) for \(2 < x < 4\).

(i) To express \(f(x) = x^2 - 4x + 7\) in the form \((x-a)^2 + b\), complete the square:

\(f(x) = (x-2)^2 - 4 + 7 = (x-2)^2 + 3\).

The range of \(f\) is \(f(x) > 3\) since \(x > 2\).

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

Rearrange to solve for \(x\):

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

Since \(x > 2\), we take the positive root:

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

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

The domain of \(f^{-1}\) is \(x > 3\).

(iii) Given \(f = hg\) and \(g(x) = x - 2\), we have:

\(f(x) = h(x)g(x) = h(x)(x-2)\).

Since \(f(x) = x^2 - 4x + 7\), equate:

\(h(x)(x-2) = x^2 - 4x + 7\).

Assume \(h(x) = x^2 + 3\) to satisfy the equation.

(i) To express \(2x^2 - 12x + 11\) in the form \(a(x + b)^2 + c\), complete the square:

\(2x^2 - 12x + 11 = 2(x^2 - 6x) + 11\)

\(= 2((x - 3)^2 - 9) + 11\)

\(= 2(x - 3)^2 - 18 + 11\)

\(= 2(x - 3)^2 - 7\)

(ii) The function \(f(x) = 2x^2 - 12x + 11\) is a quadratic function that opens upwards. It is one-one on the interval where it is either increasing or decreasing. The vertex of the parabola is at \(x = 3\). Therefore, the largest value of \(k\) for which \(f\) is one-one is 3.

(iii) To find \(f^{-1}(x)\), start with \(y = 2(x - 3)^2 - 7\).

\(y + 7 = 2(x - 3)^2\)

\(\frac{y + 7}{2} = (x - 3)^2\)

\(x - 3 = \pm \sqrt{\frac{y + 7}{2}}\)

Since \(x \leq 3\), choose the negative root:

\(x = 3 - \sqrt{\frac{y + 7}{2}}\)

Thus, \(f^{-1}(x) = 3 - \sqrt{\frac{1}{2}(x + 7)}\).

The domain of \(f^{-1}\) is \(x \geq -7\).

(iv) With \(k = 1\), \(x + 3 \leq 1\) implies \(x \leq -2\). Therefore, the largest \(p\) is -2.

The composite function \(fg(x) = f(g(x)) = f(x + 3)\).

\(f(x + 3) = 2(x + 3)^2 - 12(x + 3) + 11\)

\(= 2(x^2 + 6x + 9) - 12x - 36 + 11\)

\(= 2x^2 + 12x + 18 - 12x - 36 + 11\)

\(= 2x^2 - 7\)

(a)(i) The function \(f(x) = (x - 3)^2 - 1\) is defined for \(x < a\). The greatest possible value of \(a\) is 3, as the function is one-one up to this point.

(a)(ii) For \(a = 3\), the range of \(f\) is \(y > -1\). To find \(f^{-1}(x)\), set \(y = (x - 3)^2 - 1\), so \(y + 1 = (x - 3)^2\). Solving for \(x\), we get \(x = 3 \pm \sqrt{1 + y}\). Since \(x < 3\), we take \(x = 3 - \sqrt{1 + y}\), hence \(f^{-1}(x) = 3 - \sqrt{1 + x}\).

(b)(i) \(gg(2x) = [(2x - 3)^2 - 3]^2\). Expanding, \((2x - 3)^2 = 4x^2 - 12x + 9\). Then \(gg(2x) = (4x^2 - 12x + 6)^2\). This can be expressed as \((2x - 3)^4 - 6(2x - 3)^2 + 9\).

(b)(ii) Expanding \((2x - 3)^4 - 6(2x - 3)^2 + 9\), we get \(16x^4 - 96x^3 + 216x^2 - 216x + 81 - 24x^2 + 72x - 54 + 9\). Simplifying, \(16x^4 - 96x^3 + 192x^2 - 144x + 36\).

(i) The smallest value of \(c\) is 2. This is because the function \(f(x) = (x-2)^2 + 2\) is defined for \(x \geq c\), and the smallest value for which the function is one-to-one is when \(c = 2\).

(ii) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 + 2\). Rearrange to solve for \(x\):

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

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

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

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

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

The domain of \(f^{-1}\) is \(x > 6\) because \(f(x)\) is defined for \(x \geq 4\) and \(f(4) = 6\).

(iii) Solve \(ff(x) = 51\):

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

Set \(((x-2)^2)^2 + 2 = 51\)

\(((x-2)^2)^2 = 49\)

\((x-2)^2 = 7\)

\(x-2 = \pm \sqrt{7}\)

Since \(x \geq 4\), \(x = 2 + \sqrt{7}\).