The question is, will it recognise and use all 4gb of ram for superfetch and etc? or just 750mb of the 2gb usb?
I am using 32bit vista and i heard it can only recognise up to 2.75gb of ram
I think you may be a bit mixed up about how the 2G flash-drive works with ReadyBoost and Vista. The quick answer is that you will only have 2G of ram and faster access to your page-file (a.k.a. virtual memory). In Vista (or most anything else for that matter) the flash-drive is just another disk drive, it does not increase or change your system's available ram in any way.
ReadyBoost uses the flash-drive to 'mirror' the page-file that Vista automatically creates on your hard-drive. It then configures Vista to read page-file request from the flash drive rather than the hard-drive. This read-redirection gives Vista an improvement in overall system performance because the seek-time for a flash-drive is around 900us (its really 0 but there is some small overhead) and the seek-time for a hard-drive is around 10,000us. You will even see a more significant improvement using ReadyBoost if you are also working with other files since you do not need to actually move the disk heads to access the page-file (saving 2x seek times).
A lot of people think that you won't see any advantage with ReadyBoost unless you have a small amount of memory. This is just not true but it may be where you are coming from. ReadyBoost _really_ has nothing to do with the amount of memory in your system. It only improves the access-time for reading the page-file. Systems with little memory often access the page-file just as a 4G system that is being run hard will often access the page-file. Both can easily be put in a low-memory state, it depends on what the user is doing.