Ok, I'll give part of it a go:

For a 5K signal with 32 samples, thats 32 x 5000 = 160,000/sec times to hit the interupt. With no real numbers to use here, lets say it takes 100 instructions to "hit" the int. So thats 1,600,000 ins/sec or 1.6mips. so to start off, i would think at a min 8 meg clock (2mips)

The rest of the math depends on how the signal is generated. for instance using hpwm or maybe old school R2R D/A.

Yes i get tired of the question