Ok, you're 99% there, just a slight issue with the answer. I'll put working on how to get to where you are so it helps other people.
First you need to understand what an IP address is:
for IPv4 it's four "dotted decimal" or four octets, basically a group of four numbers between 1 - 255
In binary the address 10.0.0.1 looks like this
00001010 . 00000000 . 00000000 . 00000001
Right, so what's a subnet?
a subnet is like an IP address, a group of four numbers, separated by dots, but it does a special function.
Under the hood a binary operation is done, (AND) that separates the host digits from the network digits, this separation is described by the subnet mask
truth table for and is
0 . 0 = 0
0 . 1 = 0
1 . 1 = 1
To dumb is down and make it simple look at a single 8 bit number (90) and a subnet mask of 240 (or /4 for this small example)
0101 1010 - Address
1111 0000 - Subnet
0101 0000 - result of AND
We call the subnet address 240, because that's literally what 240 is in binary, or /4 because it starts with four 1's
So you see putting those four 1's as a mask has wiped out the last part of the number.
It allows you to show that :
0101 0000 - Address
1111 0000 - Subnet
0101 0000 - result of AND
is the range identifier, (like saying 10.0.0.0/24)
0101 0001 - Address
1111 0000 - Subnet
0101 0000 - result of AND
Is on the same network (and is the first address)
and that:
0101 1111 - Address
1111 0000 - Subnet
0101 0000 - result of AND
Is on the same network (and is the last address)
so 0101 0000 is 96 (is the network number)
so 0101 0001 is 97 (first address in range)
so 0101 1111 is 127 (last address in range)
Now Taking a look at your example.
you start with a /24 network
1111 1111 1111 1111 1111 1111 0000 0000 (24 1's followed by 8 zeros)
so you know you have 8 bits to play with, the biggest number you can make with 8 bits is 255 (and the zero)
you start with a block of 256 addresses, 0 - 255 inclusive)
you need 4 blocks.
(so 256/4 = 64, you can't use 2 addresses as those are the network address and the broacast address, but that still leaves 62 addresses available)
so the first thing is, yes, this is possible to do.
next you need to know how many bits you'd need to count to 64 addresses (0 -63) the answer is 6 (0011 1111 = 63)
so the two most significant bits of that last octet are network bits, not host bits.
thereforeyou have correctly identified the subnet mask as 26.
you only need to look at the last octet to find out the network numbers and broadcast address
so you get networks
Network 1 (last octet)
00xx xxxx (where xx can be ones or zeros)
0000 0000 - 0011 1111 = 0 - 63
Network 2 (looking only at the last octet)
01xx xxxx
0100 0000 - 0111 1111 = 64 - 127
Network 3 last octet
10xx xxxx
1000 0000 - 1011 1111 = 128 - 191
Network 4 last octet
11xx xxxx
1100 0000 - 1111 1111 = 192 - 255
I've only looked at the last octet because:
1000 1010 . 0101 0001 . 0000 1010 . 0000 0000 = 132.97.10.0
1111 1111 . 1111 1111 . 1111 1111 . 1100 0000 = 255.255.255.0
1000 1010 . 0101 0001 . 0000 1010 . 00[/i]00 0000 only these bits will change due to /26 mask)[/i]
Basically, you're almost there, 99/100!
What you're missing is that the networks that you've written down, you've (for example) listed the first address as
Subblock A: 132.97.10.0/26 – 132.97.10.63/26
You can't assign addresses that end in zero, the very first number, (whether 0, 64, 128 or 192 is the network address), not the first host address.
and the last number is the broadcast address, not an assignable host address.
e.g 10.0.0.0/24 has numbers 10.0.0.0 - 10.0.0.255
but addresses 10.0.0.1 - 10.0.0.254
.