Roughly speaking, the largeness or smallness of the number is in comparison to 1. If |x|<1, then |xn| grows smaller as n increases. Combined with the effect of the denominator in each term, the terms quickly become small, and you need to use only a few terms to get a good approximation. On the other hand, if |x|>1, then |xn| grows larger as n increases, so the terms don't get small until you've gone further in the series.
In this problem, because x<0, you end up with an alternating series. There's a theorem that says if you sum the first n terms of an alternating series to get an approximation, the error in that approximation is no bigger than magnitude of the next term in the series. (I'm being a bit sloppy here, but that's the essential concept you need.) For example, after you added the first 4 terms, you got an approximation of -0.761. The next term in the series, x4/4!, is equal to 1.285, so the theorem guarantees you that
[tex]-0.761 - 1.285 < 2^{-3.4} < -0.761 + 1.285.[/tex]To find the answer to three decimal places, you want the error to be less than 0.001, so you want to find n such that |x|n/n! < 0.001. The easiest thing to do here is try different values of n until you find one that works.
More generally, there's a formula for the remainder if you truncate the Taylor series after n terms. In a similar way, you want to find the value of n that'll make that remainder small enough for your needs. There might be some rule of thumb that let's you estimate the number of terms you need, but I'm not aware of any.