(i) To find \(f^{-1}(x)\), start with \(y = 3x - 2\). Solve for \(x\):

\(y + 2 = 3x\)

\(x = \frac{y + 2}{3}\)

Thus, \(f^{-1}(x) = \frac{x + 2}{3}\).

For \(g^{-1}(x)\), start with \(y = \frac{2x + 3}{x - 1}\). Solve for \(x\):

\(y(x - 1) = 2x + 3\)

\(yx - y = 2x + 3\)

\(yx - 2x = y + 3\)

\(x(y - 2) = y + 3\)

\(x = \frac{y + 3}{y - 2}\)

Thus, \(g^{-1}(x) = \frac{x + 3}{x - 2}\).

\(g^{-1}(x)\) is not defined for \(x = 2\).

(ii) To solve \(fg(x) = \frac{7}{3}\), first find \(fg(x)\):

\(fg(x) = f(g(x)) = 3\left(\frac{2x + 3}{x - 1}\right) - 2\)

\(= \frac{3(2x + 3)}{x - 1} - 2\)

Set \(\frac{3(2x + 3)}{x - 1} - 2 = \frac{7}{3}\)

\(\frac{3(2x + 3)}{x - 1} = \frac{7}{3} + 2\)

\(\frac{3(2x + 3)}{x - 1} = \frac{13}{3}\)

Cross-multiply to solve for \(x\):

\(3(2x + 3) = 13(x - 1)\)

\(6x + 9 = 13x - 13\)

\(6x - 13x = -13 - 9\)

\(-7x = -22\)

\(x = -8\)