the voltage resolution for an analog signal with an amplitude
1) Amplitude has no bearing on the resolution of a waveform. A pure 1KHz wave with a 2V amplitude has the same resolution as a 4V amplitude wave has the same resolution of a 2uV wave. The source wave has "infinite" resolution. I use that in quotes because all real signals have a noise floor which limits their effective resolution.

2) Your DAC (is DAQ a typo?) has a resolution of (Vdd-Vee)/(2^bit depth). That is to say your DAC is limited in how small a change in the signal wave it can actually see. The equation I gave you is the generic form of finding the smallest change in voltage that your DAC can properly distinguish. Your DAC is also limited by the voltage rails it uses. If the signal is larger than the voltage rails than the DAC can't "see" that part of the wave.

3) Nyquist theory states that you must sample MORE THAN two times the frequency of a wave to be able to recover it without aliasing. I.E. using a 1KHz wave as the highest frequency you'd like to recover, you must sample MORE THAN 2000 times per second (>2KHz) to recover the waveform. I'm emphasizing the "MORE THAN" as I see quite a few people who mistakenly think that sampling at exactly twice the rate will recover the signal. If you sample at exactly twice the frequency of the wave you want to capture, you will alias a DC signal.
That section of your question confuses me, as DAC's can't READ. They take a digital value and convert it to an analog voltage (or current). However, I've provided and example of how and ADC would react based on sampling rate.

Edit 1: In the problem you were given the maximum sampling rate. Using math inequalities you would simply reverse the example to find your answer. I.E. if the ADC could only sample at 2KHz, then you can only "see" frequencies recorded that were LESS THAN 1KHz.