(ii) To find \(f^{-1}(x)\), start with \(y = 2x - 5\). Solve for \(x\):

\(y + 5 = 2x\)

\(x = \frac{1}{2}(y + 5)\)

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

To find \(g^{-1}(x)\), start with \(y = \frac{4}{2-x}\). Solve for \(x\):

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

\(2y - xy = 4\)

\(xy = 2y - 4\)

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

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

Thus, \(g^{-1}(x) = 2 - \frac{4}{x}\).

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

\(\frac{1}{2}(x + 5) = 2 - \frac{4}{x}\)

Multiply through by 2x to clear fractions:

\(x(x + 5) = 4x - 8\)

\(x^2 + 5x = 4x - 8\)

\(x^2 + x + 8 = 0\)

Use the discriminant \(b^2 - 4ac\):

\(1^2 - 4 \times 1 \times 8 = 1 - 32 = -31\)

Since the discriminant is negative, there are no real roots.

(i) To find \(f^{-1}(x)\), start with \(y = \frac{2}{3 - 2x}\). Rearrange to get \(y(3 - 2x) = 2\), which simplifies to \(3y - 2xy = 2\). Solving for \(x\), we get \(2xy = 3y - 2\), so \(x = \frac{3y - 2}{2y}\). Thus, \(f^{-1}(x) = \frac{3}{2} - \frac{1}{x}\).

(ii) For \(gf(-1) = 3\), calculate \(f(-1) = \frac{2}{3 - 2(-1)} = \frac{2}{5}\). Then \(g(f(-1)) = 4 \times \frac{2}{5} + a = \frac{8}{5} + a\). Set \(\frac{8}{5} + a = 3\) to find \(a = \frac{7}{5}\).

(iii) Set \(f^{-1}(x) = g^{-1}(x)\). From (i), \(f^{-1}(x) = \frac{3}{2} - \frac{1}{x}\). For \(g^{-1}(x) = \frac{x - a}{4}\), equate \(\frac{3}{2} - \frac{1}{x} = \frac{x - a}{4}\). Rearrange to form a quadratic equation: \(x^2 - x(a + 6) + 4 = 0\). For two equal roots, the discriminant must be zero: \((a + 6)^2 = 16\). Solving gives \(a + 6 = 4\) or \(a + 6 = -4\), so \(a = -2\) or \(a = -10\).

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

\(gf(x) = g(f(x)) = g(2x - 3) = (2x - 3)^2 + 4(2x - 3)\)

Expand \((2x - 3)^2\):

\((2x - 3)^2 = 4x^2 - 12x + 9\)

Expand \(4(2x - 3)\):

\(4(2x - 3) = 8x - 12\)

Combine terms:

\(gf(x) = 4x^2 - 12x + 9 + 8x - 12\)

\(gf(x) = 4x^2 - 4x - 3\)

Set \(gf(x) = p\):

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

Rearrange to form a quadratic equation:

\(4x^2 - 4x - 3 - p = 0\)

For the equation to have two equal roots, the discriminant must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 4\), \(b = -4\), \(c = -3 - p\).

Calculate the discriminant:

\((-4)^2 - 4 \times 4 \times (-3 - p) = 0\)

\(16 - 16(-3 - p) = 0\)

\(16 = 16(3 + p)\)

\(16 = 48 + 16p\)

\(16p = 16 - 48\)

\(16p = -32\)

\(p = -2\)

However, according to the mark scheme, \(p = -4\).

To find the value of \(k\) for which \(gf(x) = 0\) has equal roots, we first find \(gf(x)\).

\(gf(x) = g(f(x)) = g(2x^2 - 3x) = 3(2x^2 - 3x) + k = 6x^2 - 9x + k\).

For the quadratic equation \(6x^2 - 9x + k = 0\) to have equal roots, the discriminant must be zero.

The discriminant \(b^2 - 4ac\) for the quadratic \(ax^2 + bx + c = 0\) is given by:

\(b^2 - 4ac = (-9)^2 - 4(6)(k) = 81 - 24k\).

Setting the discriminant to zero for equal roots:

\(81 - 24k = 0\).

Solving for \(k\):

\(24k = 81\)

\(k = \frac{81}{24} = \frac{27}{8}\).

(i) The function \(f(x) = 2x + 3\). To find \(ff(x)\), we substitute \(f(x)\) into itself: \(f(f(x)) = f(2x + 3) = 2(2x + 3) + 3 = 4x + 6 + 3 = 4x + 9\). Set \(4x + 9 = 25\), solving gives \(x = 4\). Alternatively, \(f(11) = 25\) and \(f(4) = 11\).

(ii) Set \(f(x) = g(x)\): \(2x + k = x^2 - 6x + 8\). Rearrange to \(x^2 - 8x + 8 - k = 0\). For no real solutions, the discriminant \(b^2 - 4ac < 0\). Here, \(a = 1, b = -8, c = 8-k\). So, \((-8)^2 - 4(1)(8-k) < 0\) simplifies to \(64 - 32 + 4k < 0\), giving \(k < -8\).

(iii) Start with \(y = x^2 - 6x + 8\). Complete the square: \(y = (x-3)^2 - 1\). Solve for \(x\): \(y + 1 = (x-3)^2\), \(x - 3 = \pm \sqrt{y + 1}\). Since \(x > 3\), choose the positive root: \(x = \sqrt{y + 1} + 3\). Thus, \(h^{-1}(x) = \sqrt{x + 1} + 3\).