(i) To find the set of values of \(x\) for which \(f(x) > 15\), solve the inequality:

\(x^2 - 2x > 15\)

Rearrange to form a quadratic equation:

\(x^2 - 2x - 15 = 0\)

Factorize the quadratic:

\((x - 5)(x + 3) = 0\)

The solutions are \(x = 5\) and \(x = -3\).

Test intervals around the roots to find where \(f(x) > 15\):

For \(x < -3\), choose \(x = -4\):

\(f(-4) = 16 + 8 = 24 > 15\)

For \(x > 5\), choose \(x = 6\):

\(f(6) = 36 - 12 = 24 > 15\)

Thus, \(x < -3\) and \(x > 5\).

(ii) To find the range of \(f\), complete the square:

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

The minimum value of \((x-1)^2\) is 0, occurring at \(x = 1\).

Thus, the minimum value of \(f(x)\) is \(-1\).

Therefore, the range of \(f\) is \(y \geq -1\).

Since \(f\) is a quadratic function and not one-to-one, it does not have an inverse.

(a) The function \(f(x) = -3x^2 + 2\) is a downward-opening parabola. The vertex occurs at \(x = 0\), but since \(x \leq -1\), the maximum value of \(f(x)\) is at \(x = -1\). Calculating \(f(-1) = -3(-1)^2 + 2 = -1\). Therefore, the range of \(f\) is \(y \leq -1\).

(b) To find the inverse, start with \(y = -3x^2 + 2\). Rearrange to solve for \(x\):

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

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

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

Since \(x \leq -1\), we take the negative root:

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

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

(i) To express \(2x^2 + 8x - 10\) in the form \(a(x + b)^2 + c\), complete the square:

\(2x^2 + 8x - 10 = 2(x^2 + 4x) - 10\)

\(= 2((x+2)^2 - 4) - 10\)

\(= 2(x+2)^2 - 8 - 10\)

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

Thus, \(a = 2\), \(b = 2\), \(c = -18\).

(ii) The least value of \(y\) occurs at the vertex of the parabola, which is at \(x = -2\). Substituting \(x = -2\) into the equation:

\(y = 2(-2)^2 + 8(-2) - 10 = 8 - 16 - 10 = -18\)

Thus, the least value of \(y\) is \(-18\) when \(x = -2\).

(iii) To find the set of values of \(x\) for which \(y \geq 14\), solve:

\(2x^2 + 8x - 10 \geq 14\)

\(2x^2 + 8x - 24 \geq 0\)

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

\((x+6)(x-2) \geq 0\)

The solution is \(x \geq 2\) or \(x \leq -6\).

(iv) For \(f\) to be one-one, the domain must be restricted to where the function is either increasing or decreasing. The vertex is at \(x = -2\), so the smallest \(k\) is \(-2\).

(v) To find \(f^{-1}(x)\), solve \(y = 2x^2 + 8x - 10\) for \(x\):

\(y = 2(x+2)^2 - 18\)

\(y + 18 = 2(x+2)^2\)

\(\frac{y+18}{2} = (x+2)^2\)

\(x+2 = \pm \sqrt{\frac{y+18}{2}}\)

Since \(x \geq -2\), choose the positive root:

\(x = \sqrt{\frac{y+18}{2}} - 2\)

Thus, \(f^{-1}(x) = \sqrt{\frac{x+18}{2}} - 2\).

(a) To express \(f(x) = -2x^2 - 8x - 13\) in the form \(-2(x + a)^2 + b\), complete the square:

\(f(x) = -2(x^2 + 4x) - 13\)

\(= -2((x + 2)^2 - 4) - 13\)

\(= -2(x + 2)^2 + 8 - 13\)

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

Thus, \(a = 2\) and \(b = -5\).

(b) The vertex form \(-2(x + 2)^2 - 5\) shows the maximum value is \(-5\) when \(x = -2\). Since \(x < -3\), \(f(x) < -7\).

(c) To find \(f^{-1}(x)\), start with \(y = -2(x + 2)^2 - 5\).

Rearrange to solve for \(x\):

\(y + 5 = -2(x + 2)^2\)

\((x + 2)^2 = -\frac{y + 5}{2}\)

\(x + 2 = \pm \sqrt{-\frac{y + 5}{2}}\)

\(x = -2 \pm \sqrt{-\frac{y + 5}{2}}\)

Thus, \(f^{-1}(x) = -2 - \sqrt{\frac{x+5}{2}}\) or \(f^{-1}(x) = -2 + \sqrt{\frac{x+5}{2}}\).

To find \(f^{-1}(x)\), start by setting \(y = f(x) = 2 - \frac{3}{4x - p}\).

Rearrange to solve for \(x\):

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

\((y - 2)(4x - p) = -3\)

\(4xy - 8x = py - 2p - 3\)

\(4x(y - 2) = p(y - 2) - 3\)

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

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

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

Thus, \(f^{-1}(x) = \frac{p}{4} - \frac{3}{4x - 8}\).

For \(f^{-1}(x) \equiv f(x)\), equate the expressions:

\(2 - \frac{3}{4x - p} = \frac{p}{4} - \frac{3}{4x - 8}\)

Equating the coefficients gives \(p = 8\).