To create 4 subnets, 2 bits must be borrowed from the host portion (2^2 = 4).

A /28 subnet has 2^(32-28) = 2^4 = 16 total addresses. Subtracting 2 for the network and broadcast addresses leaves 14 usable addresses.

The network address is obtained by performing a bitwise AND between the IP address and the subnet mask: 192.168.1.10 AND 255.255.255.0 = 192.168.1.0.

In a /16 subnet, the valid host range is from 172.16.0.1 to 172.16.255.254, with 172.16.0.0 as the network address and 172.16.255.255 as the broadcast address.

A subnet mask of 255.255.255.128 corresponds to /25, which allows for 2^(32-25) = 2^7 = 128 total addresses. Subtracting 2 for the network and broadcast addresses leaves 126 usable hosts.

The subnet mask 255.255.255.192 corresponds to /26, which allows for 2^(26-24) = 4 subnets from a Class C network.

The subnet mask 255.255.255.248 corresponds to /29, as it has 29 bits set to 1.

A /22 subnet has 2^(32-22) = 2^10 = 1024 total addresses. Subtracting 2 for the network and broadcast addresses leaves 1022 usable host addresses.

To support 500 hosts, a subnet mask of 255.255.254.0 (/23) is required, providing 510 usable addresses.

To accommodate 30 usable addresses, a /27 subnet mask (255.255.255.224) is needed, providing 32 total addresses (30 usable).

To accommodate at least 500 usable addresses, a subnet must have at least 512 total addresses, which corresponds to a /23 subnet mask (255.255.254.0).

In the subnet 10.0.0.0/8, the valid host addresses range from 10.0.0.1 to 10.255.255.254. Therefore, 10.0.1.1 is a valid host address.

A subnet mask of 255.255.255.240 (/28) provides 16 total addresses, allowing for 14 usable IP addresses.