Answer: (a)(i) \(12\), (a)(ii) \(72\), (b) \(4200\).

Count the choices in stages, applying any restrictions before multiplying the number of possibilities together.

(a)(i) If no girls are next to each other, the only pattern is

\(\text{girl, boy, girl, boy, girl}\).

The girls can be arranged in \(3!\) ways and the boys in \(2!\) ways, giving \(3!\times2!=12\).

(a)(ii) With no restriction, there are \(5!\) arrangements. If the two boys are next to each other, treat them as one block. Then there are \(4!\times2!\) such arrangements.

Therefore the number with the boys not next to each other is

\(5!-4!\times2!=120-48=72.\)

(b) Put Anjie in a group first. Choose the other 3 people in Anjie's group from the 10 people who are neither Anjie nor Bubay:

\({}^{10}C_3\).

Then choose 3 people to be in Bubay's group from the remaining 7 people:

\({}^{7}C_3\).

The last 4 people form the final group. Hence the number of ways is

\({}^{10}C_3\cdot{}^7C_3=120\cdot35=4200.\)

This gives the required answer: (a)(i) \(12\), (a)(ii) \(72\), (b) \(4200\).