(i) Let the first term be \(a = 5\frac{1}{3} = \frac{16}{3}\) and the common ratio be \(r\). The fourth term is given by \(ar^3 = 2\frac{1}{4} = \frac{9}{4}\).

Thus, \(\frac{9}{4} = \frac{16}{3}r^3\).

Solving for \(r^3\), we have:

\(r^3 = \frac{9}{4} \times \frac{3}{16} = \frac{27}{64}\)

Taking the cube root, \(r = \frac{3}{4}\) or 0.75.

(ii) The sum to infinity of a geometric progression is given by \(S_\infty = \frac{a}{1-r}\).

Substituting the values, \(S_\infty = \frac{5\frac{1}{3}}{1 - \frac{3}{4}} = \frac{16}{3} \times 4 = \frac{64}{3}\).

Thus, \(S_\infty = 21\frac{1}{3}\) or 21.3.