解：
3: 求2^m的3余数
2^0, r = 1
2^1, r = 2
2^2, r = 1
循环。所以
2^(2n), r = 1。
2^1000=2^(2(500)), r = 1
被3除之后的余数=1

5: 求2^m的5余数
2^0, r = 1
2^1, r = 2
2^2, r = 4
2^3, r = 3
2^4, r = 1
循环。所以
2^(4n), r = 1。
2^1000=2^(4(250)), r = 1
被5除之后的余数=1

7: 求2^m的7余数
2^0, r = 1
2^1, r = 2
2^2, r = 4
2^3, r = 1
循环。所以
2^(3n), r = 1。
2^999=2^(3(333)), r = 1
2^1000, r = 2
被7除之后的余数=2

11: 求2^m的11余数
2^0, r = 1
2^1, r = 2
2^2, r = 4
2^3, r = 8
...
2^10, r = 1
循环。所以
2^(10n), r = 1
2^1000 = 2^(10(100)), r = 1
被11除之后的余数=1

13: 求2^m的13余数
2^(12n), r = 1
2^(12(83))=2^996, r = 1
2^997, r = 2
2^998, r = 4
2^999, r = 8
2^1000, r = 3
被13除之后的余数=3

17: 求2^m的17余数
2^(8n), r = 1
2^1000 = 2^(8(125)), r = 1
被17除之后的余数=1

19: 求2^m的19余数
2^(18n), r = 1
2^(18(55))= 2^990, r = 1
2^1000, r = 17
被19除之后的余数=17