No, Problem 1 is just about integers.

The number of stops is the number of times the train stops after you get on it. In other words, the station where you get on doesn't count, but the station where you get off does. (This is standard terminology used by people who ride buses and trains.)

For example, suppose A, B, C, and D are consecutive stations on a line. Then a ride from A to B is 1 stop and would cost 1 token. A ride from A to C would cost 2 tokens (1 stop from A to B, one more stop from B to C), and a ride from A to D would cost 3 tokens.

If C, E, F are consecutive stations on a different line and you go from A to F by switching at C, then your ride would cost 4 tokens: 2 stops from A to C, 2 stops from C to F, and no cost for switching lines.

In 2b, a pair of subway lines can intersect any nonzero number of times, but exactly one of these intersections is a station.

For the purposes of this problem, to show that a particular set of points form the vertices of the required polyhedron, you need to specify twelve 5-tuples of points that correspond to the faces of your polyhedron and then prove that each of these 5-tuples does indeed form a "crescent" as defined in the problem. (Technically, before you conclude that the resulting shape is a polyhedron, there is a lot more that would need to be verified -- e.g. each edge belongs to exactly two faces, each pair of intersecting faces meets only at a vertex or an edge, etc. But you don't need to prove any of those on the Quiz.)

We are more interested in how you structure your argument than in the details of your calculations. So if you have a lot of similar calculations, you can just give the details for one and say that the others are analogous. And definitely feel free to appeal to symmetry when appropriate.

If you're using any of these tools, then you've probably already figured out the coordinates of the vertices. So, as explained in the previous answer (see above), what remains is to specify which 5-tuples of vertices form faces and to prove that these faces are crescents. However, since the coordinates of the vertices involve irrational numbers, any computations on a calculator, Desmos, or most other software would be only approximate, hence insufficient for a proof. In this sense, a 3D model in Desmos is basically equivalent to a paper model: it's great for visualization and checking your intuition, and if you've made it, you should include it in your solution. But because it's only an approximation, it's not a proof.

Yes, you can, as long as your net still remains a single connected piece. (I.e. you should be able to go from any face to any other face in the net via the uncut edges.)

Yes.

No, the road does not have to be straight. It just has to cross from one plot to another through edges, not vertices. This condition is meant to ensure that all the plots in a single territory are connected via edges ("orthogonal adjacency"). So you can't have a territory that consists of two pieces joined only by a single vertex: to get from one of those pieces to the other, the road would have to pass through the vertex, which is not allowed.

The law only requires that each individual fence be finite; it does not say anything about the total length of all the fences.

The statement in 5b is true. If you think you have a counterexample, it must violate one of the problem conditions. Reread the problem carefully!

Yes, non-integer times do count.