What does it mean to say 'vectors are independent of the co-ordinate system' and why is this useful?
It's related to the Three Rules of Linear Algebra:
1) Don't pick a basis.
2) DON'T pick a basis.
3) If you do pick a basis, choose very carefully.
The point is, vectors are not lists of numbers. The list of numbers, together with a choice of basis, does determine a vector. But change the basis and your numbers change. Much of the stuff you need to do with linear algebra does not depend one bit on the basis.
For instance, you can compute the rank of a matrix by doing row-echelon reduction. Change the basis, and you get what looks like an entirely different matrix, but if you compute its rank, it's the same. That's because the two matrices represent the same linear transformation, under two different bases. And the rank is really a property of the transformation, so it makes sense that it's the same for every representation of that transformation.
Likewise, vectors have properties that are independent of the chosen basis. This question is trying to get you to think about vectors as objects in their own right, and to consider what properties they have that are independent of your choice of basis.
it's like this: space and forces exist, regardless of how we coordinatize them. the polar coordinates of a vector don't turn it into a "different" vector than its cartesian coordinates, those are just names we invented to describe the vectors. you can make up your own names, if its useful for you, your names just need to be "comprehensive" (this is what spanning is all about), and "unambiguous" (this is what linear independence is all about).
or consider this: you have a function which represents some actual physical relationship. this function just does what it does, it doesn't suddenly turn into a tiger if you change from a fourier series to a taylor series. the coefficients of a series representation don't "define it" they just "describe" it.
when we pick physical units for measuring something, we're choosing a basis (this doesn't seem so confusing with a one-dimensional vector space, hmm?)