As Fer said to get 5 subnet u use a /27 with the no of valid subnets as 8 and a block size of 32 with 30 being the valid hosts ( remember one goes for n/w address and one for broadcast ) so now.. calculate the range
Valid subnets --> 0,32,64,96,128,160, etc etc.. till 224
so to answer ur 1.3) how to solve it? the usuable IPs which u will get is from 192.168.100.65 - 192.168.100.94 ( 192.168.100.64 as the n/w address and 192.168.100.95 as the broadcast address) .. and that makes your answer right
Note: You still got 3 more usuable subnets which u can have fun with
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