(i) Consider each family as a single unit or 'block'. There are 4 families, so we arrange these 4 blocks in different orders:

\(4! = 24 ext{ ways}\)

Within each family, the members can be arranged among themselves. The number of arrangements for each family is:

2! ext{ for the Lin family,}

3! ext{ for the O’Connor family,}

5! ext{ for the Ahmed family,}

2! ext{ for the Burton family.}

Thus, the total number of arrangements is:

\(4! imes 2! imes 3! imes 5! imes 2! = 829440\)

(ii) First, arrange the 8 children. There are:

8! ext{ ways to arrange the children.}

To ensure no two adults are consecutive, place the 6 adults in the 9 gaps created by the 8 children (before the first child, between children, and after the last child). Choose 6 out of these 9 gaps:

9 imes 8 imes 7 imes 6 imes 5 imes 4 ext{ ways to choose and arrange the gaps.}

Thus, the total number of arrangements is:

\(8! imes 9 imes 8 imes 7 imes 6 imes 5 imes 4 = 2438553600\)