(i) Consider the group of boys as one unit and the group of girls as another unit. There are 2 units to arrange, which can be done in \(2!\) ways. Within the boys' group, the 5 boys can be arranged in \(5!\) ways, and within the girls' group, the 6 girls can be arranged in \(6!\) ways. Therefore, the total number of arrangements is \(5! \times 6! \times 2 = 172800\).

(ii) Arrange the 6 girls first, which can be done in \(6!\) ways. This creates 7 gaps (before, between, and after the girls) where the boys can be placed. Choose 5 out of these 7 gaps to place the boys, which can be done in \(\binom{7}{5}\) ways. The 5 boys can be arranged in these gaps in \(5!\) ways. Therefore, the total number of arrangements is \(6! \times \binom{7}{5} \times 5! = 6! \times 7 \times 6 \times 5 \times 4 \times 3 = 1814400\).