Hi Hig,
Subnetting eh... I feel your pain mate.
Let me start from the beginning as, like many things, there's more than one way to skin a cat and subnetting is definitely no exception.
Let's get some binary maths out of the way - here are all the powers of 2 for upto 8 bits (we won't cover subnetting across octet boundaries just yet!):
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
So, let's say we start with your your 172.16.0.0 network which, being a Class B network, has the default subnet mask 255.255.0.0.
If we want to split it into 8 subnets we need to borrow 3 bits from the host portion as per the table above. If we wanted 9 subnets we'd have to go up to the next bit boundary, i.e. 4 bits (giving us 16 subnets).
The require mask is worked out by setting the first 3 bits to 1, and setting the rest to 0:
128 64 32 16 8 4 2 1
1 1 1 0 0 0 0 0 = 128 + 64 + 32 = 224
(forgive the non-fixed-width font)
Hence our mask is 255.255.224.0
To work out what subnets that gives us we need to work out the 'block size'. The easiest way to do that is simply to do 256-mask i.e. 256-224 = 32 hence our subnets, starting from 0 (the so-called 'subnet zero' is generally allowable nowadays), are:
172.16.0.0 255.255.224.0 (or 172.16.0.0/27)
172.16.32.0/27
172.16.64.0/27
172.16.96.0/27
172.16.128.0/27
172.16.160.0/27
172.16.192.0/27
172.16.224.0/27
Taking any one subnet, say 172.16.64.0/27, we can work out how many addresses this subnet has in it, and the range...
Remember we started off with 16 host bits, but borrowed 3 leaving 13 host bits for us to use. 2^13 = 8192 addresses. The range is 172.16.64.0 - 172.16.95.255 (one less than the next subnet) but we cannot use the first one as this refers to the network, nor the last as this is the broadcast address. The true number is therefore (2^13) - 2 = 8190 and the range 172.16.64.1 - 172.16.95.254.
Make any sense?!
Mathew
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He said it has always been given to him so he never had to sort it lol!
(God help me when I start supernetting 
Let me read thorugh it again lol 
I always wonderd what the subnet was and it still don't know after a reading that. Good luck with it hig.