Answer with explanation:
→Number of Integers from 1 to 100
=100(50 Odd +50 Even)
→50 Even =2,4,6,8,10,12,14,16,...............................100
→50 Odd=1,3,,5,7,9,..................................99.
→Sum of Two even integers is even.
→Sum of two odd Integers is odd.
→Sum of an Odd and even Integer is Odd.
(a)
Number of ways of Selecting 2 integers from 50 Integers ,so that their sum is even,
=Selecting 2 Even integers from 50 Even Integers , and Selecting 2 Odd integers from 50 Odd integers ,as Order of arrangement is not Important, ,

=4900 ways
(b)
Number of ways of Selecting 2 integers from 100 Integers ,so that their sum is Odd,
=Selecting 1 even integer from 50 Integers, and 1 Odd integer from 50 Odd integers, as Order of arrangement is not Important,

=2500 ways