First, arrange the five older children. There are 5! ways to do this.

Next, consider the gaps between and around these five children where the three youngest can be placed. There are 6 such gaps.

Choose 3 out of these 6 gaps to place the youngest children. This can be done in \(\binom{6}{3} = 20\) ways.

Arrange the three youngest children in these chosen gaps. There are 3! ways to arrange them.

Thus, the total number of ways is \(5! \times \binom{6}{3} \times 3! = 120 \times 20 \times 6 = 14400\).

(b) To find the number of different seating arrangements for the 8 friends, consider the following:

Rajid and Sue must sit on the same side of table X. There are 2 ways to choose which side of table X they sit on. For each choice, there are 2 ways to arrange Rajid and Sue. The remaining 6 friends can be seated in any of the remaining 6 seats (4 at table Y and 2 on the opposite side of table X). This gives:

\(2 \times 2 \times 6! = 2880\)

(c) To find the number of different arrangements if Rajid and Sue stand next to each other, but neither is at an end of the line:

Consider Rajid and Sue as a single unit. This unit can be placed in any of the 6 positions in the line (positions 2 to 7, since neither can be at an end). For each position, there are 2 ways to arrange Rajid and Sue within the unit. The remaining 6 friends can be arranged in the remaining 6 positions. This gives:

\(6 \times 2 \times 6! = 7200\)

(ii) First, arrange the 7 cup cakes, which can be done in \(7!\) ways. This creates 8 gaps (including the ends) where the brownies can be placed. Choose 4 out of these 8 gaps to place the brownies, which can be done in \(^8P_4\) ways. Therefore, the total number of arrangements is \(7! \times ^8P_4 = 8467200\).

(iii) Treat the 4 chocolate biscuits as a single unit. This gives us 7 units to arrange: 1 chocolate unit, 6 shortbread biscuits, and 2 gingerbread biscuits. The total arrangements of these units is \(\frac{9!}{6! \times 2!}\), accounting for the identical shortbread and gingerbread biscuits. Therefore, the total number of arrangements is 252.

(i) To find the number of different lists Hannah can make, we use permutations since the order matters. The number of ways to choose and arrange 5 singers from 15 is given by the permutation formula:

\(^{15}P_5 = \frac{15!}{(15-5)!} = \frac{15!}{10!}\)

Calculating this gives:

\(15 \times 14 \times 13 \times 12 \times 11 = 360360\)

Thus, there are 360360 different lists.

(ii) For the specific order of performers (male, female, male, female, male), we calculate the number of ways to choose each performer:

- First performer (male): 5 choices

- Second performer (female): 10 choices

- Third performer (male): 4 choices (since one male is already chosen)

- Fourth performer (female): 9 choices (since one female is already chosen)

- Fifth performer (male): 3 choices (since two males are already chosen)

The total number of lists is:

\(5 \times 10 \times 4 \times 9 \times 3 = 5400\)

Thus, there are 5400 such lists.

Since Mark is already seated in the front of taxi P, we have 7 friends left to arrange.

Jon and Sarah are sitting in the back of taxi P, so we treat them as a single unit initially. This leaves us with 5 units to arrange: Jon-Sarah, 1 more person in the back of P, and 4 people in taxi Q.

First, choose 1 more person to sit in the back of taxi P from the remaining 5 friends: \\(\binom{5}{1} = 5 \\\).

Now, arrange Jon and Sarah in 2 ways (Jon-Sarah or Sarah-Jon): \\(2! = 2 \\\).

Arrange the 4 people in taxi Q: \\(4! = 24 \\\).

The total number of arrangements is: \\(5 \times 2 \times 24 = 240 \\\).

Since there are 2 possible ways to arrange Jon and Sarah, multiply by 2: \\(240 \times 2 = 480 \\\).