These are all good questions
So, I'm guessing, the resultant focal length of the system (whatever that is), will also depend upon the distance between the lenses. How does the power of the lens fit into all this?
That is correct. In the limit where your lenses are thin the powers of two lenses is added as follows:
$$\phi_\text{tot} = \phi_1 + \phi_2 - \phi_1\phi_2\tau$$
where the individual powers are given by $\phi_1 = 1/f_1,\phi_2 = 1/f_2$, and $\tau = t/n$. $t$ is the spacing between the two lens elements and $n$ is the index of refraction of the medium between your two lenses ($n=1$ in vacuum). The effective focal length is then $f_\text{effective} = 1/\phi_\text{tot}$
There is a distinction between the effective focal length $f_\text{effective}$, the rear focal length $f_\text{R}$, and the front focal length $f_\text{F}$.
$$f_\text{R} = n_\text{R}f_\text{effective}\quad \quad f_\text{Front} = -n_\text{Front}f_\text{effective}$$
I.e. the front focal length is measured from the front principal plane to the front focal point (negative distance for a positive effective focal length) and the rear focal length is measured from the rear principal plane to the rear focal point (positive distance for a positive effective focal length).
Also, what do they mean by the 'resultant power of the lenses' in the first place? The resultant system of lenses is not going to be like these lenses which have negligible thickness. Where are you going to measure the focal length from?
This is a very important point. For a system of multiple lenses you measure the front and rear focal lengths from what are called the front and rear principal planes (rather than measuring directly from a lens element). For the case of a single lens the front and rear principal planes are located at the lens. For two lenses, the front principal plane is shifted from the first lens element a distance
$$d_\text{F} = + \frac{\phi_1}{\phi_\text{tot}} \frac{n_\text{F}\;t}{n}$$
the rear principal plane is shifted from the second lens element a distance.
$$d_\text{R} = -\frac{\phi_1}{\phi_\text{tot}}\frac{n_\text{R}\;t}{n}$$
For the above two equations I am defining the first lens to be on the left and the second lens on the right. The separation is $t$ and the index of the medium between the two lenses is $n$. The index of the medium to the right of the two lenses is $n_\text{R}$ and the index of the medium to the left of the two lenses is $n_\text{F}$. A value of $d_\text{F}$ or $d_\text{R}$ less than zero indicates a shift of the respective principal plane to the left; a value greater than zero indicates a shift to the right.
Disclaimer: this response is only a very brief introduction to the ideas of gaussian optics and the cardinal points. This stuff can be really confusing for anybody, especially when you consider all of the sign conventions. Also, these equations are really only valid in the paraxial limit for rotationally symmetric systems. Having said that, these basic formulations can be expanded to a system of any number of lenses -- not just 2. If you really want to understand this stuff there are surely some good books on this material. Hecht seems to be the book for intro optics, although I haven't read him. Check the table of contents to make sure gaussian optics is covered (it should be) because that's quite an expensive text.
The exit pupil of any optical system is defined as the image of the controlling aperture stop (AS) formed by the subsequent optical components. The controlling aperture stop is the optical component which limits the maximum cone of rays from an object point which can actually be processed by the whole system (basically it controls how much light from each point actually makes it all the way through). In a system like a camera we deliberately set the AS by having an iris which opens and closes, but in the absence of an iris one of the lenses limits the maximum angle. In the case of a simple telescope this is generally the main front lens, and the exit pupil is its image formed by the eyepiece lens as your formula states. Though if you put too small an eyepiece lens then this could be the AS itself in which case it will also be the exit pupil.
In general the exit pupil is designed to form at the position where the eye will be placed and further that it will be the same size as the pupil of the eye itself, this ensures that the maximum possible amount of light can be transmitted from the object to the viewer and that the image fills the eye. The exit pupil limits the cone angle of rays illuminating each point of the image.
The aperture stop should not be confused with the field stop which controls the field of view of the system. This aperture limits the maximum angular distance an object can be from the optical axis and still be seen. Its image in the subsequent components is called the Exit Window and effectively defines the total size of the final image. To an observer on the image side this window appears to limit the area of the of the image.
Best Answer
Yes, the lenses have to be separated by $f_1+f_2$. To see why, consider in the ray optics picture what happens when the distance is different and the input rays are parallel instead of originating at a point in the focus. Or consider in the Fourier optics picture where the Fourier transform plane would be in such a system.