Keep in mind that seating arrangements at a circular table are invariant with respect to rotation. Thus, once we seat the first person, all permutations are relative to where that person sits.
(a) Let's sit the girls first. One girl sits down. There are $3!$ ways to seat the other girls. Once the girls sit down, there are $4!$ ways to arrange the boys, giving
$$3!4! = 6 \cdot 24 = 144$$
possible seating arrangements in which the boys and girls alternate seats.
Alternate solution: There are $4!$ ways of sitting the girls. However, there are four rotations of the girls that do not change their relative order. Thus, there are
$$\frac{4!}{4} = 3!$$
distinguishable ways of sitting the girls. Once they are seated, there are $4!$ ways of sitting the boys. Thus, there are $4!3! = 144$ ways of sitting the girls in which the boys and girls alternate seats.
(b) We seat the boy and girl in adjacent seats first. Once she sits down, he can sit either to her right or to her left, giving two ways of arranging the boy and girl who must sit in adjacent seats. Relative to her, there are $3!$ ways of seating the remaining girls. Once they are seated, there are $3!$ ways of seating the remaining boys, giving
$$2! \cdot 3! \cdot 3! = 2 \cdot 6 \cdot 6 = 72$$
seating arrangements in which the boys and girls sit in alternate seats and a particular boy and girl sit in adjacent seats.
Alternate solution: We showed above that we can seat the girls in $3!$ distinguishable ways. There are $2$ ways we can sit the particular boy next to the particular girl (on her right or her left). Once he has been seated, there are $3!$ ways to seat the remaining boys, which yields
$$3! \cdot 2 \cdot 3! = 6 \cdot 2 \cdot 6 = 72$$
seating arrangements in which a particular boy sits next to a particular girl.
(c) We subtract the number of ways we can sit a particular boy and girl in adjacent seats when the boys and girls sit in alternate seats from the total number of ways they can sit in alternate seats, which yields
$$3!4! - 2!3!3! = 144 - 72 = 72$$
seating arrangements in which the boys and girls sit in alternate seats and a particular boy and girl do not sit in adjacent seats.
First seat the $4$ girls around the table, which can be done in $(4-1)!$ ways (why?, Check "circular permutation"). Now in between these girls, there are $4$ gaps, where you would have to seat the $5$ boys, two of whom must sit together. So we club the two boys to form a single unit and see that the $4$ units can be seated in the $4$ gaps in $4!$ ways. However the two boys who form the unit, can be rearranged among themselves in $2$ ways. So total number of ways is $2!\times3!\times4!$.
Best Answer
If we imagine the seats numbered from $1$ through $10$ around the table, so that they can each be individually identified even if we rotate the whole table, your answer is almost correct. The girls can be seated either in the even-numbered seats or in the odd-numbered seats ($2$ possibilities); in either case they can be seated in $5!$ different orders; and the boys can then be assigned to the remaining $5$ seats in $5!$ different ways, so we get a total of $2\cdot5!\cdot5!$ possibilities.
However, in problems about seating people around a circular table you’re generally supposed to assume that the seats are not individually identifiable, so that two seatings that arrange the participants in the same cyclic order are considered the same seating. A seating that puts Anna in seat $1$ is considered the same as a seating that puts her in seat $2,3$, or indeed any other seat, provided that each of the $10$ people still has the same left and right neighbors. Thus, each single seating in this sense corresponds to $10$ seatings if the seats are numbered, and we therefore have to divide by $10$ to get the correct number:
$$\frac{2\cdot5!\cdot5!}{10}=\frac{5!\cdot5!}5=4!\cdot5!\;.$$
This result can be obtained in another way as well. There are $4!$ different cyclic listings of the $5$ girls (because listings $ABCDE$ and $DEABC$ are the same cyclically), so we pick one of them and seat the girls in alternate seats around the table in that order. There are then $5!$ ways to assign the boys to the $5$ remaining seats, for a total of $4!\cdot5!$ possible arrangements. The assumption that only the cyclic order matters, not the assignment to specific seats, is taken care of at the first step in this approach.