(i) The formula for the sum to infinity of a geometric progression is \(\frac{a}{1-r} = 256\), where \(a = 64\). Solving for \(r\):

\(\frac{64}{1-r} = 256\)

\(1-r = \frac{64}{256}\)

\(1-r = \frac{1}{4}\)

\(r = \frac{3}{4}\)

(ii) The formula for the sum of the first \(n\) terms is \(S_n = a \frac{1-r^n}{1-r}\). For the first ten terms:

\(S_{10} = 64 \frac{1-(0.75)^{10}}{1-0.75}\)

\(S_{10} = 64 \frac{1-0.0563}{0.25}\)

\(S_{10} = 64 \times 3.7748\)

\(S_{10} = 242\)