(i) To find \(f(x)\), we start with the inverse function \(f^{-1}(x) = \frac{1 - 5x}{2x}\). We need to solve for \(x\) in terms of \(y\):

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

\(2xy = 1 - 5x\)

\(2xy + 5x = 1\)

\(x(2y + 5) = 1\)

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

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

The domain of \(f\) is determined by the range of \(f^{-1}\), which is \(x > -\frac{9}{4}\).

(ii) To find \(f^{-1}g(x)\), substitute \(g(x) = \frac{1}{x}\) into \(f^{-1}\):

\(f^{-1}\left(\frac{1}{x}\right) = \frac{1 - 5\left(\frac{1}{x}\right)}{2\left(\frac{1}{x}\right)}\)

\(= \frac{x - 5}{2}\)

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