(i) Let the probability that a random integer is less than or equal to 4 be p. Since there are 4 integers (1, 2, 3, 4) less than or equal to 4 out of 9, p = 4/9. The probability that at least 2 of the 5 integers are less than or equal to 4 is given by:

\(P(\text{at least 2}) = 1 - P(0, 1)\)

\(= 1 - \left(\frac{5}{9}\right)^5 - \binom{5}{1} \left(\frac{4}{9}\right) \left(\frac{5}{9}\right)^4\)

\(= 0.735\)

(ii) We know that the mean \(np = 96\) and the variance \(npq = 32\), where \(q = 1 - p\).

From \(np = 96\), we have \(p = \frac{96}{n}\).

From \(npq = 32\), we have \(np(1-p) = 32\).

Substituting \(p = \frac{96}{n}\) into \(npq = 32\), we get:

\(n \left(\frac{96}{n}\right) \left(1 - \frac{96}{n}\right) = 32\)

\(96 \left(1 - \frac{96}{n}\right) = 32\)

\(96 - \frac{9216}{n} = 32\)

\(\frac{9216}{n} = 64\)

\(n = 144\)

Since \(p = \frac{96}{144} = \frac{2}{3}\), \(k\) is such that \(P(\leq k) = \frac{2}{3}\). Therefore, \(k = 6\).