You can get an injection from any collection of disjoint figure-eights on the plane to Q4 by picking any rational point in the interior of each of the two circles making up the figure, so there can only be countably many such disjoint figures. This is an injection because if another figure-eight had the same rational points as its representative, then it’s easy to show that the two figures must intersect somewhere.
I forgot about this one. For the first problem:
You can get an injection from any collection of disjoint figure-eights on the plane to Q4 by picking any rational point in the interior of each of the two circles making up the figure, so there can only be countably many such disjoint figures. This is an injection because if another figure-eight had the same rational points as its representative, then it’s easy to show that the two figures must intersect somewhere.