Well, if there are two computers on the network, with IPs of 192.168.28.175 and 192.168.15.12, only the 192.168 is common of both IP addresses, so a suitable network ID would be 192.168. I wouldn't suggest putting 192.168.0.0 down - it's just about correct, but not the perfect answer. If, for example, the two IPs were 192.168.28.175 and 192.168.28.176, a suitable network ID would be 192.168.28, as this is common to both addreses.
Part two, well we've already established that the network ID is 192.168, so other computers on the network can have any valid IP address that starts with 192.168 (except for 192.168.0.0, and also strictly 192.168.0.1 and 192.168.255.255 too). So this would be the range available - 192.168.0.1 to 192.168.255.254.
Hope this helps a bit. Good luck in the exam.