(i) Set \(f(x) = g(x)\):

\(k - x = \frac{9}{x+2}\)

Multiply through by \(x+2\):

\((k-x)(x+2) = 9\)

\(kx + 2k - x^2 - 2x = 9\)

Rearrange to form a quadratic:

\(x^2 + (2-k)x + 9 - 2k = 0\)

For two equal roots, the discriminant must be zero:

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

\((2-k)^2 - 4 \times 1 \times (9-2k) = 0\)

Solve for \(k\):

\(k = 4\) or \(k = -8\)

For \(k = 4\), the root is \(x = 1\).

For \(k = -8\), the root is \(x = -5\).

(ii) Solve \(fg(x) = 5\) when \(k = 6\):

\((6-x) \cdot \frac{9}{x+2} = 5\)

\(54 - 9x = 5x + 10\)

\(44 = 14x\)

\(x = 7\)

(iii) Express \(g^{-1}(x)\) in terms of \(x\):

Let \(y = \frac{9}{x+2}\)

\(x = \frac{9}{y} - 2\)

\(g^{-1}(x) = \frac{9}{x} - 2\)