Given subnet : 128.119.40.128/26
we have 26 n/w bits and 6 host bits.
To assign any IP address then that IP must be present in this subnet.
128.119.40.10000000 Bold indicates host bits.
We can assign any address between 000001 to 111110.
That gives 128.119.40.129/26 to 128.119.40.190/26.
2)Now ISP owns 128.119.40.64/26
to get 4 subnets we need 2 bits.
we borrow 2 bits from host id part.remaining bits in host id are 4.
basic idea is we will get the form of a.b.c.d/28.
With 4 bits we get 2^4 = 16 hosts per subnet including n/w and broadcast addresses.
1st subnet: 128.119.40.64/28
2nd subnet: 128.119.40.80/28. (64 + 16)
3rd subnet:128.119.40.96/28 (80 + 16)
4th subnet: 128.119.40.112/28. (96+16).