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.
A lens does not have have one specific magnification, it depends on the positioning of the lens. When neglecting aberrations, the workings of a lens can be simplified with the following equation,
$$
\frac{1}{f} = \frac{1}{v} + \frac{1}{b},
$$
where $f$ is the focal length of the lens, $v$ is the distance from the object to the lens and $b$ the distance from the lens to the image of the object. This is demonstrated in the image below, including three principal rays (these only apply for thin lenses).
Here $\text{F}_1$ and $\text{F}_2$ are the two focal points of the lens, with $f_1$ and $f_2$ as their respective focal lengths (these are often equal to each other, which is also assumed in the first equation).
The resulting magnification, $M$, will be equal to the ratio between $h_1$ and $h_2$, which when expressed in terms of $v$ and $b$ looks as follows,
$$
M = \frac{v}{b}.
$$
When you have a lens with a given focal length then you have two equations with three unknown. So, when you want to calculate the magnification you would not have a unique solution. However manufactures probably want to add a label to their lenses which a layman can understand. For this they probably will use eyepiece magnification,
$$
M_e = \frac{250\ mm}{f}
$$
where the numerator is equal to the least distance of distinct vision, which is roughly 250 mm for a human with normal vision.
However if you do not know the focal length you can use the following equation,
$$
\frac{1}{f} = (n - 1) \left[\frac{1}{R_1} - \frac{1}{R_2} + \frac{(n - 1) d}{n R_1 R_2}\right],
$$
where $n$ is the refractive index of the lens material, $d$ the thickness of the lens, $R_1$ and $R_2$ the radius of curvature of the two sides of the lens.
Best Answer
John, your method for determining approximate lens focal lengths is a valid one, although since your eye will automatically accommodate to some extent, accuracy will be limited.
The apparent discrepancies follow from the rather obscure convention of labeling lens "Power" in units of Diopters, which are inverse meters.
The power P of an optic is defined as the inverse of the focal length f in meters:
$$ P = \frac{1}{f} $$
Thus a lens of power $-3$ has a focal length of $-1/3$ meter or $-33$ $cm$, not far from the $f= -30$ $cm$ you estimated. Similarly, a power of $-0.25$ Diopters corresponds to $-4$ $m$ or $-400$ $cm$, a very weak negative lens which would scarcely be noticeable when combined with a fairly high powered positive lens (magnifier).