This started as a comment, but outgrew the space.
Edit1, 3(a) vs Edit2, (3) on their face sound like different questions to me: (forevery knot is there a polyhedron), vs. (is there a polyhedron such that forevery knot) for which the knot is a billiard trajectory.
From a solution to Edit1, 3(a), you can get a convex set that works as in Edit2,3, but I don't see why a polyhedron would be implied. If 3(a) is true, then take a countable sequence of polyhedra together with billiard trajectories that cover all possible knot types. Take a sequence of points on $\mathbb {RP}^2$ with disjoint neighborhoods of radius $\epsilon_i$. Now compress the $i$th polyhedron by an affine map until the points of contact with the knot are nearly on a pair of parallel planes with tangent planes nearly parallel to that plane, and graft this in to $S^2$ near the pair of antipodal points representing the $i$th point chosen in $\mathbb{RP}^2$.
There is an implication that forevery knot there is a convex set ... implies forevey knot there is a polyhedron ... . A convex set that works for a knot, or finite collection of knots, can be modified to a polyhedron: just use the tangent planes at the points of contact to delineate a polyhedron.
My guess is that: 3(a) is true. I'm imagining taking a picture of a knot as a plait, and then constructing a convex tube just for it, where the knot mostly bounces up and down toward you and away from you, but bends and crosses where necesary. If I get a concrete construction, I'll fill in specifics --- currently it's just an idea. I would also guess that no single polyhedron will work for all knots. It's an interesting challenge to try to describe special properties of knots associated with a specific polyhedron.