Problem 2.
192.168.10.0 /26 (IP Address)
So Subnet Mask is 255.255.255.192 (i.e. /26 ) ( 255 means 8 bits ON and 192 means 2 bits ON )
(8+8+8+2 == 26)
192 means 2 ON bit and 6 OFF
Valid Subnets = 256-192 ==>64
Block :: 0 , 64 , 128, 192
Network = 2^2==>4
host = 2^6-2==> 62