(1) There are $A_{10}^{10}$ ways to arrange the boys and $A_{10}^{10}$ ways to arrange the girls. Considering the relative positions between the two lines, we know there are
$$
A_{10}^{10} \cdot A_{10}^{10} \cdot A_{2}^{2}=(10!)^{2} \cdot 2!=2 \cdot(10!)^{2}
$$

different cases.
(2) It is only necessary to place 20 people in 20 different positions, so there are $A_{20}^{20}=20$! different cases.
(3) Consider the pairing of boys and girls one by one: there are 10 possible partners for boy A; 9 possible partners for boy B; ...; and 1 possible partner for boy J. Additionally, there is the issue of the positions between the two lines, so there are $2 \cdot 10$! different cases.
【Example 9】There are 10 boys and 10 girls lining up in two rows to dance together, (1) if boys and girls stand in separate rows; (2) if they stand in two rows without considering gender, with 10 people in each row; (3) if boys and girls stand in separate rows, not considering the order within each row, only considering which two people are paired. How many different situations are there in each case?