a. Assuming here that subnet zero is allowed. 32 subnets needs 5 bits as 2^5 = 32. So you need to borrow 5 bits from the fourth octet. So the subnet mask will be /24+5 = /29.
b. There are 3 bits remaining from the 8 bits of the fourth octet. 3 bits gives you 2^3-2=6 hosts in each subnet.
c. If subnet zero is counted in this question. Then network address of subnet 1 will be 126.96.36.199. So The first address is 188.8.131.52+1 = 184.108.40.206 and the last address is 220.127.116.11+7-1 = 18.104.22.168