(i) The probability of not getting a 3 on a single die is \(\frac{5}{6}\). Therefore, the probability of not getting a 3 on all 9 dice is \(\left( \frac{5}{6} \right)^9\). The probability of getting at least one 3 is:

\(1 - \left( \frac{5}{6} \right)^9 = 1 - 0.1938 = 0.806\)

(ii) The probability of getting at least one 3 when \(n\) dice are thrown is \(1 - \left( \frac{5}{6} \right)^n\). We need this probability to be greater than 0.9:

\(1 - \left( \frac{5}{6} \right)^n > 0.9\)

\(\left( \frac{5}{6} \right)^n < 0.1\)

Taking logarithms on both sides:

\(n \log \left( \frac{5}{6} \right) < \log(0.1)\)

\(n > \frac{\log(0.1)}{\log \left( \frac{5}{6} \right)}\)

Calculating gives \(n > 12.6\), so the smallest integer \(n\) is 13.