(i) For \(f(x) = \frac{3}{2x+1}\), as \(x \to 0^+\), \(f(x) \to 3\). As \(x \to \infty\), \(f(x) \to 0\). Thus, the range of \(f\) is \(0 < f(x) < 3\).

For \(g(x) = \frac{1}{x} + 2\), as \(x \to 0^+\), \(g(x) \to \infty\). As \(x \to \infty\), \(g(x) \to 2\). Thus, the range of \(g\) is \(g(x) > 2\).

(ii) \(fg(x) = f(g(x)) = \frac{3}{2\left(\frac{1}{x} + 2\right) + 1} = \frac{3}{\frac{2}{x} + 4 + 1} = \frac{3}{\frac{2 + 5x}{x}} = \frac{3x}{2 + 5x}\).

(iii) Let \(y = \frac{3x}{2 + 5x}\). Then \(y(2 + 5x) = 3x\).

Rearranging gives \(2y + 5xy = 3x\) or \(3x - 5xy = 2y\).

Solving for \(x\), we get \(x(3 - 5y) = 2y\) and \(x = \frac{2y}{3 - 5y}\).

Thus, \((fg)^{-1}(x) = \frac{2x}{3 - 5x}\).