(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.