(i) To find the probability that the numbers on four dice add up to 5, consider the combinations: (1, 1, 1, 2), (1, 1, 2, 1), (1, 2, 1, 1), (2, 1, 1, 1). Each die has a probability of \(\frac{1}{6}\) for a specific number. The probability for one combination is \(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{1296}\). There are 4 such combinations, so the total probability is \(4 \times \frac{1}{1296} = \frac{1}{324}\).

(ii) Let \(p = \frac{1}{324}\) be the probability from part (i) and \(q = 1 - p = \frac{323}{324}\). We use the binomial probability formula for exactly 1 or 2 successes in 7 trials: \(P(1,2) = \binom{7}{1} p^1 q^6 + \binom{7}{2} p^2 q^5\). Calculate each term: \(\binom{7}{1} \times \frac{1}{324} \times \left(\frac{323}{324}\right)^6 + \binom{7}{2} \times \left(\frac{1}{324}\right)^2 \times \left(\frac{323}{324}\right)^5 = 0.0214\).