(i) To find \(ff(x)\), we substitute \(f(x)\) into itself:

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

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

Simplifying the numerator:

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

Simplifying the denominator:

\(2\left(\frac{x+3}{2x-1}\right) - 1 = \frac{2(x+3)}{2x-1} - 1 = \frac{2x+6 - (2x-1)}{2x-1} = \frac{7}{2x-1}\)

Thus, \(ff(x) = \frac{7x}{7} = x\).

(ii) To find \(f^{-1}(x)\), we set \(y = \frac{x+3}{2x-1}\) and solve for \(x\):

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

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

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

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

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