(i) To find \(gg(x)\), substitute \(g(x) = 2x - 3\) into itself:

\(gg(x) = g(g(x)) = g(2x - 3) = 2(2x - 3) - 3 = 4x - 6 - 3 = 4x - 9\).

(ii) To find \(f^{-1}(x)\), start with \(y = \frac{1}{x^2 - 9}\).

Rearrange to express \(x\) in terms of \(y\):

\(y = \frac{1}{x^2 - 9} \Rightarrow x^2 - 9 = \frac{1}{y} \Rightarrow x^2 = \frac{1}{y} + 9\).

Taking the square root, \(x = \sqrt{\frac{1}{y} + 9}\).

Thus, \(f^{-1}(x) = \sqrt{\frac{1}{x} + 9}\).

The domain of \(f^{-1}\) is \(x > 0\) because \(y > 0\) in the original function.

(iii) Solve \(fg(x) = \frac{1}{7}\):

\(fg(x) = \frac{1}{(2x - 3)^2 - 9} = \frac{1}{7}\).

\((2x - 3)^2 - 9 = 7 \Rightarrow (2x - 3)^2 = 16\).

\(2x - 3 = \pm 4 \Rightarrow 2x = 7 \text{ or } 2x = -1 \Rightarrow x = \frac{7}{2} \text{ or } x = -\frac{1}{2}\).

Since \(x > 3\), the solution is \(x = \frac{7}{2}\).