B:
[tex]1.55780644887275 ms \times 1000000 = 1.56E6 ms[/tex]
[tex]1.56\times 10^6 ms \times \frac{1 s}{1000 ms} = 1.56\times 10^3 s[/tex]

C:
This is the part I am not sure about.
My logic is that if the last digit is ±3, that is 14 decimal places. So I took:
[tex]3\times 10^{-14} \times 1000000[/tex]
That gives 3e-8 ms
I wanted to simplify this a bit, so I converted it to seconds:
[tex]\frac{3\times 10^{-8}}{1000} = 3\times 10^{-11}[/tex]
Simplifying it to simpler units:
[tex]3\times 10^{-11} \times 1\times 10^12 = 30 ps[/tex]

Therefore my answer is ±30 ps.

But seriously, the book never mentions uncertainty, so I did the first thing that was logical to me. It seems to have the tendency to ask about things it doesn't discuss, though so far it has only been various common formulas.

Well the well-known formula for the uncertainty on a function f(x1,x2,...xn), where each x has a known uncertainty is:
[tex]\sigma^2_f = \sum_{i=1}^{i=n}(\frac{\partial f}{\partial x_i})^2\sigma^2_{x_i}[/tex]

I'm glad you are so confident. I am not. I should note that this is an introductory physics course in college, usually taken by freshmen, and is the first chapter of the book, so I doubt such an equation would be used yet?

all it is is that one rotation's time has the given uncertainty. if that rotation is repeated 1000000 times, wouldn't the uncertainty be multiplied by the same? perhaps not, but the equation seems rather complicated for the problem.

It's a good equation, but I don't know what sigma is, nor F, not to mention I don't have a function to which to apply it...

Well the equation is supposed to be completely general. Sigma represents uncertainties, and f is the function on which you know the uncertainty. Here in this case, sigma = 3 * 10^-14s, this is the uncertainty on the variable P (P for period). f is given by f = 1000000P. In this case, yes, the the derived uncertainty is just 1000000 times the uncertainty on P (as one can check, if they wish, by using the formula I gave). In this case, the formula need not be applied and using your intuition is fine.

However, suppose you wanted to find the uncertainty on your answer for part b. To get the correct answer, you would have to use the formula I gave.

I think it is a worthwile formula to be familiar with, especially if you plan on later doing any labs in physics.

Thanks. Well, do I notice correctly there is a partial derivative in there, assuming \partial means partial derivative? I've only gone through calc 1 so far, and we didn't cover anything like that. When I can, I'd love to figure out how to calculate this, though.

Calculating partial derivatives is really very simple. You don't need any knowledge passed Cal1 to do so. Simply differentiate with respect to one variable, while holding the others constant. For example, let [itex]f(x,y) = x^2y[/itex]. Then
[tex]\frac{\partial f}{\partial x} = 2xy[/tex]
and
[tex]\frac{\partial f}{\partial y} = x^2[/tex].