You should always find an answer that is a formula, and then only apply significant figures once you get to the one final step of substituting your numbers back into the problem in place of variables.
Avoid multiple intermediate steps of substituting numbers at all costs. Not only will this save your pencil a lot of work, but it will also cause your answer to be more accurate, as rounding errors can pile up, even when using a calculator.
I encounter significant digits much more often in chemistry than in physics. So basically:
Say you have a ruler with centimeter and millimeter markings. You measure the length of a pencil, and it comes out to somewhere in between 8.6 cm and 8.7 cm. It seems a touch closer to 8.6 than to 8.7. So, you say that the pencil is 8.63 cm long. The last digit implies that it is $\pm.01$. This way, the value could be 8.62, 8.63, 8.64, or anywhere in between. The most that you know is that it is definitely closer to 8.6 than 8.7, and the range from 8.62-8.64 just about covers your uncertainty about the measurement.
If you wanted to be absolutely precise, every single measurement you make and quantity you calculate would have a tolerance based on the limitations of your measuring apparatus. Of course, it would be cumbersome to keep writing $\pm.01$ every time, so it is simply assumed that the value is known exactly except for the last digit, which is uncertain.
Now when you do calculations, you can't use the value that you found, because it has some uncertainty associated with it. To be correct, you would have to carry out multiple calculations, first on the lower bound, and then on the upper bound, to figure out what the uncertainties of your new quantity are. This doubles the number of calculations you need to make, and is just cumbersome and tedious. That is why the rules for significant digits arose. They are a guideline for figuring out what kind of uncertainty your new quantity has without having to make any extraneous calculations. Thus, when you multiply 2 numbers, one with 3 sig figs and the other with 2, you will know for sure that the product will have 2 sig figs, one of which is absolutely certain, the other slightly uncertain.
To reiterate, in the example above, our value of 8.63 cm for the length of the pencil has 3 significant digits; two of which are absolutely certain (8 and 6), and the last one certain to $\pm1$.
Best Answer
It depends on the type of number. If the dimensionless number arises from pure math or counting, then it essentially has infinite precision and doesn't limit the number of digits you report. This is the case for the $2$ relating radius and diameter, or the $1/2$ in the kinetic energy formula, or the $N!$ you often see in statistical mechanics.
$\pi$ is the same, but if you only plug in finitely many digits into your calculator, your are limited by that. Thus $3.14$ has $3$ significant figures, even if you are approximating the exact value of of $\pi$.
Other dimensionless numbers are indeed measured, and thus have limited precision as well. For example, the fine-structure constant or the g-factor related to the electron gyromagnetic ratio are all measured and have uncertainties.
In general the question of having units vs. being dimensionless is orthogonal to the question of being exact vs. having uncertainty.