Part I: Lapse Rates
For this explanation, adiabatic lapse rates are negative.

There are two different ways to assess stability:
1. You use the temperature profile of a layer and determine its lapse rate. Then you compare this environmental lapse rate (ELR) to the relevant adiabatic lapse rate (ALR). If ELR > ALR, the layer is stable, if ELR = ALR it's neutral, and if ELR < ALR it's unstable.
2. You consider a single parcel of air and decide whether an incremental lifting from its original position would make it return to the original position (stable), stay where the deviation put it (neutral) or continue rising (unstable). A parcel rises, when its temperatur is higher than the surrounding air, and sinks when its temperature is lower (ignoring the effect of humidity).

So much for Questions 1 + 2. If this is not enough, study the book (e.g. about orographic precipitation).

Question 3:
(a) Even if we say we want the temperature as a function of elevation, we want the elevation as a function of temperature, since we are grounded in the meteorological tradition. Sorry about that.
(c) You take a parcel from a certain height (i.e. with a specific starting temperature) and force it to rise. It will cool adiabatically, at the DALR. If you take the parcel from the surface (-31 F) and lift it to 6,100 ft, it has a temperature of -65 F. It is much colder than its surroundings and therefore will sink again. In this sinking and warming process it no sooner reaches the Temperature of the surrounding air than at the surface. It has returned to its original position, and can be considered stable.

Question 4: 45 degrees Celsius.

Question 5:
Is explained in the book as far as pollution under the circumstances of an inversion is concerned.
You have to make sure that the plume of smoke rises and mixes, i.e. is unstable. If you assume that the smoke has the temperature of the surrounding air it will definitely not rise and mix  if you build the smokestack lower than the inversion, i.e. at least 500 ft high (this is the minimum height the question asks for). If it is dry, it will still be stable at all heights given, so that you don't really know what to do.
If you know the temperature of the smoke, you can tell at every height whether it will be stable there or not. For example, if it is 32 F at the surface, it will rise from there, since it is warmer. If it is unsturated, it will cool 0.55 F per 100 ft, so that it has a temperature of 29.25 F at 500 ft and 28.7 F at 600 ft.Therefore, the air will stop rising and start spreading out somewhere between  500 ft and 600 ft. If it starts at 33 F at the surface, it will be 29.7 F at 600 ft, and continue to rise. If it is warm enough, it will actually be able to break through the inversion and spread out in greater heights, without a smokestack at all. Actually, smoke from an incinerator is neither completely dry nor cold.

Question 6: Use the Ideal Gas Law.
pV=MTR.
It tells you that the product of pressure p and volume V is proportional to the product of mass M and temperature T (with the proportionality constant R). Now, what happens in the experiment?

First, the water and the air in the can are heated, i.e. T goes up. According to the equation, either p or V have to go up, or M has to go down. Since the can is open, p is the same inside the can and outside of it. However, V increases (we saw molecules leaving the can). Since the molecules are gone, M is decreased.

Now, we turn the can around. By putting the mouth under water, we prevent gas molecules from entering or leaving the can in the gas state. However, some of the water vapor molecules leave the gas phase which is oversaturated (because the water surface it is in contact with has lower temperature than the old one in the heated can). So, M goes down even further. In addition, since the heat source is removed, T goes down, too. That means, that p and/or V have to go down to make up for that. In fact, they both do: p first goes down (since the partial pressure of the water vapor is gone). The fluid water body and the thin walls of the can admit some kind of adjustment of the volume to the high pressure gradient: Water is sucked in, and the walls implode, so that the gas volume in the can is reduced. Some people correctly mentioned, that no complete equilibrium between the pressures in the collapsed can and the atmosphere is reached since the gravitation of the water prevents it....