(ii) To arrange the 9 mugs such that the 3 china mugs are separated, first arrange the 6 plastic mugs. There are \(6!\) ways to arrange these. This creates 7 gaps (before, between, and after the plastic mugs) to place the china mugs. Choose 3 out of these 7 gaps, which can be done in \(\binom{7}{3}\) ways. Arrange the 3 china mugs in these gaps in \(3!\) ways. Therefore, the total number of arrangements is \(6! \times \binom{7}{3} \times 3! = 720 \times 35 \times 6 = 151200\).

(iii) Treat the 3 identical red mugs as a single unit. This gives us 12 units to arrange: 1 red unit, 4 blue mugs, and 7 yellow mugs. The total number of arrangements is \(\frac{12!}{4! \times 7!}\), accounting for the identical blue and yellow mugs. Therefore, the number of arrangements is \(\frac{479001600}{24 \times 5040} = 3960\).