1: The law is even, even, even, even, even, even, even, even, even. . . . . 50÷3= 16+2, and the last two digits are odd numbers, so there are 16×2+2=34 odd numbers.

2. Suppose there are m students in each group and each student has n trees, then (3m+ 1)×n=364, which are substituted into 1, 2, 3 and 4 in turn. . .

Get 1, 91; 4,28; 9, 13; 17,7; 30,4; 12 1, 1

The total * * * is three students in these six situations, 9 1 tree; Or 12 students with 28 trees; Or 27 students, 13 trees; Or 5 1 student, 7 trees; Or 90 students, 4 trees; Still 363 students, a tree.

According to practical and reasonable suggestions, take the third or fourth case, that is, 27 students, each planting 13 trees; Or 5 1 student, planting 7 trees each.

This problem is not detailed enough, so it is better to have a picture.