(i) We know that 225 out of 900 cartons contain more than 1002 millilitres. Therefore, the probability \(P(X > 1002) = \frac{225}{900} = 0.25\).

Using the standard normal distribution, \(P(Z > z) = 0.25\) corresponds to \(z = 0.674\) (using standard normal tables or calculator).

Standardizing, we have:

\(\frac{1002 - \mu}{8} = 0.674\)

Solving for \(\mu\):

\(1002 - \mu = 0.674 \times 8\)

\(1002 - \mu = 5.392\)

\(\mu = 1002 - 5.392 = 996.608\)

Rounding to the nearest whole number, \(\mu = 997\).

(ii) The probability that a carton contains more than 1002 millilitres is \(p = 0.25\). We want the probability that exactly 2 out of 3 chosen cartons contain more than 1002 millilitres.

This is a binomial probability problem with \(n = 3\), \(k = 2\), and \(p = 0.25\).

\(P(2) = \binom{3}{2} \times (0.25)^2 \times (0.75)^1\)

\(P(2) = 3 \times 0.0625 \times 0.75\)

\(P(2) = 0.140625\)

Rounding to three significant figures, the probability is 0.140.