Op-amp and manipulating its linear output range
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