Assuming I filled in the table correctly, the question wants me to use it in order to help me figure out what the difference between significant digits and precision is. I am not sure what I am looking for. I know precision is dependent on the last significant digit but not on the other significant digits. Is there a pattern I am missing?
You have filled out the table correctly.
I think the point of the exercise is to see that precision and significance are two different concepts. precision is the smallest difference that is reliably measured between two values. For example at work, most of the scales we use have a precision of $0.02$ pounds (guess where I live). Thus if you weigh two items and they are reported as being the same weight, in reality, they could actually differ by nearly $0.02$ lbs. For example, if one truly weighs $0.9901$ lbs, and the other truly weighs $1.0099$ lbs, both will still be reported as $1.00$ lbs by the scale, as that is the nearly multiple to $0.02$ to both of them. The absolute error (or just error) in the scales is half the precision, or $0.01$. If the weight reported is $W$, then I know that the real weight of the part differs from $W$ by at most $0.01$. It could be anywhere from $W - 0.01$ to $W + 0.01$. Any further away, and I would have gotten a different value from the scale.
But consider this: I weigh two parts. For one the weight is $0.02$ lbs. For the other, the weight is $100.00$ lbs. Now I go figure out how much weight these two parts add to an airplane. As it happens, there is only one of the hundred pound part used, but there are $5000$ of the tiny $0.02$ lb fasteners. $5000 \times 0.02 = 100$ lbs as well. The total weights are the same.
Or are they? I actually know that the weight of the heavy part is between $99.99$ and $100.01$ lbs, and that is the limits of how much weight it can be contributing to the aircraft. I know that the weight of the tiny part is somewhere between $0.01$ and $0.03$ lbs. So $5000$ of them will weigh anywhere between $50$ and $150$ lbs, a very significant variance.
This is what significant digits measures. The $100.00$ lb weight I measured for the heavy part has 5 significants. I know 4 of the digits exactly. The 5th one I know to $\pm2$ (so really, its more like 4.8 significant digits, but we usually don't make that sort of distinction). The maximum the weight I have could be wrong is only $0.01\over 100.00$, or $0.01\%$ of the total weight.
On the other hand, for the tiny part, I only know 1 significant digit, and that one to only $\pm2$. The maximum this weight could be wrong is a whopping $0.01\over0.02$, or $50\%$ of the total weight. These ratios are called the relative error. As you can see, it is the relative error, not the absolute error, that is important when mixing measurements that differ vastly in size.
So precision is a measure of absolute error, which is generally useful when dealing with items that are all roughly the same size. But significant digits is a measure of relative error, which is more useful when mixing measurements that are of vastly different sizes.