I run Cubase 64-bit and WaveLab 7 on Windows 7 64-bit, with an RME 64-bit (only) driver and use ASIO without any problems. 32-bit applications run fine on 64-bit Windows as long as you have 64-bit drivers for all your hardware. I’ve run WaveLab (albeit version 6) in as little as 512MB RAM without any problem, so I don’t think 64-bit technology has anything to offer as far as WaveLab is concerned at this point, and moreover, if there were a 64-bit version of WaveLab 7, I don’t think anyone would be able to tell the difference subjectively between 32- and 64-bit.
As usual the confusion here is the use of the term “64-bit” in relation to both CPU architecture and audio resolution.
Referring to CPU architecture, 64-bit is almost always a good thing as it allows the use of practically unlimited RAM, and even assists 32-bit applications running under a 64-bit OS; so for example, Cubase users who use large sample libraries which need to loaded into RAM will certainly benefit from the use of both a 64-bit OS and the 64-bit version of the application. Even the 32-bit version of Cubase can benefit from being run in a 64-bit OS, something which many of us are doing (usually in cases where there’s no 64-bit version of a favourite VST plugin, and the VSTBridge doesn’t work).
In the case of audio resolution, when audio is passed between VST plugins, it is done only after each sample is converted to what’s called a “floating point” number – think of this as mapping e.g. a 16-bit (integer) sample as it might exist in an audio file to a value between 0 and 1, but with all possible values between 0 and 1 available, thus giving infinitely more resolution during processing. Only when the result is converted back to integer samples (e.g. 16-bit, 24-bit) for playback via an audio interface or for burning to disk is the resolution decreased.
So, in summary, as far as audio processing is concerned, it does not matter whether WaveLab 7 as an application is compliled for 32-bit or 64-bit CPU architecture, the audio processing would be identical because this happens in the floating-point domain.