Electronic – Total output resistance (Rout) is calculated when Early voltage is completely unspecified and neglected. How

Well the input range is clearly 150mV and the output range is found in this sentence: -

without exceeding the typical guaranteed linear output range

This means that for a normal op-amp (that can swing to within 1.5 volts of the power rails) on a "normal" supply of +/-15 volts, the output swing will be 30 volts - 3 volts = 27 volts swing (p-p).

This means the gain will be 27 volts/150 mV = 180.

Now you have to decide whether the power rails are differently applied in your question and whether the op-amps to be used can swing to much closer to the power rails that +/- 1.5 volts.

The input being offset is easy to deal with. Firstly I note that the question doesn't imply non-inverting gain so, the non-inverting input can be set at -25 mV. This is halfway between -100mV and +50mV.

In other words the +50mV input is 75mV above the -25 mV bias point and the 100 mV input is 75mV below the -25 mV bias point i.e. input symettry about -25mV.

For a +50mV input (the actual input relative to -25mV is 75 mV) and, with a gain of -180, will drive the output 13.5 volts negative.

When the input is -100mV the real input is -75mV and this will drive the output 13.5 volts positive.

Everywhere in between will be linear as one would expect from an op-amp.

One small point is the -25mV at the positive input - this will centre the output at -25 mV when there is no input present and you could argue that it would tend to force the output (for a +50mV input) to be -13.525 volts i.e. slightly into clipping. This could be countered by calculating the gain to get exactly -13.5 volts.

If you want to double check all of this try using a free simulator like LTSpice

2. Source is not bypassed by a source bypass capacitor

Yet you clearly have the source bypassed in your diagram A. So that'd get you that question wrong.

I see the issue you have. If you assume gate leakage is zero, then you can't possibly solve for what your professor asks. However, you aren't provided the gate leakage current. Gate leakage is practically impossible to guess. So, you are right to be confused.