Yes, if you use limited-precision data types. But you have it the wrong way, if you first cancel out the $BIGNUM (ie. reorder to $BIGNUM - $BIGNUM + 1) the answer is 1; if you first evaluate $BIGNUM+1, the answer is 0 because $BIGNUM+1 has no representation distinct from $BIGNUM. Limited-precision arithmetic is not, in general, associative. Still arithmetic, though, just not in the ring of integers. But the whole point of the article was that it's, of course, possible to do better and get exact results.
no, because only in our imaginations and in no place in the universe can we ignore significance of measurements. If we are sending a spaceship to an interstellar object 1 light year away from earth, and the spaceship is currently 25 miles from earth (on the way), you are insisting that you know more about the distance from earth to the object than you do if you think that that distance from the spaceship to the galaxy is 587862819274.1 miles
i) Answer is 0 if you cancel out two expression (10^100)
ii) Answer is 1 if you compute 10^100 and then add 1 which is insignificant.
How do you even cater for these scenarios? This needs more than arithmetic.