1. First, through enumeration, three people may sit in positions 2, 4 and 6, 2, 4 and 7, 2, 5 and 7, 3, 5 and 7 respectively. There are four possibilities. If three people sit casually, it is 3P3, 4*3P3=24.

2. In the gap insertion method, four dance programs have five gaps, but there can be no programs at the beginning and end. Four dance programs are randomly arranged in 4P4. If all five vacancies are filled, the six singing programs will be divided into 1, 1, 1, 1, 2. First, randomly select two of the six programs and put them in. If only four gaps are inserted, six singing programs can be divided into 1, 1, 1, 3 or 1, 1, 2, 2. If it is 1, 1, 1, 3, 3. If only three gaps are inserted, they can be divided into 1, 1, 4 or 1, 2,3 or 2,2,2, (6C4*5P3+6P3*3P2+6P2*4P2)*4P4.

4p 4[6p 2 * 5p 5+2(6p 3+6p 2 * 4p 2)5p 4+(6 C4 * 5p 3+6p 3 * 3p 2+6p 2 * 4p 2)* 4p 4]= 399 1680

3.3 There are four ways to put letters into 1 mailbox; Put two mailboxes, two 3C2, two letters together, two letters in four mailboxes 4P2 and three mailboxes 4P3.

4+3C2*4P2+4P3=64