Solution:

a.
Assign coordinate axes making a $120^{\circ}$ angle so that the center of the hexagon is at $(0,0)$, the rightmost frog is at $(1,0)$, and the top left frog is at $(0,1)$. Then, the remaining four frogs are at $(1,1)$, $(-1,0)$, $(0,-1)$, and $(-1,-1)$. At each jump, if two frogs' coordinates differ by $(x, y)$, then the jumping frog moves $(3x, 3y)$. That is, each coordinate changes by a multiple of three. However, the goal has both coordinates divisible by three, and none of the frogs start with both coordinates divisible by three, so it cannot be done.

b.
One solution method is to repeat the coordinate method of the previous problem, but now encountering fractions when we use Rule 2. Since the jumping frog now moves $(3x/2, 3y/2)$ when the frogs are separated by $(x, y)$, looking at the first coordinate a frog's jump will now be
$$\frac{p}{q} \rightarrow \frac{p}{q} + \frac{3qx}{2q} = \frac{2p + 3qx}{2q}$$
If $p$ is not divisible by $3$ before this jump, the numerator of this new fraction is still not divisible by $3$. Since the frogs start with at least one coordinate's numerator not divisible by $3$, the frogs can never reach a location where both coordinates have numerators that are divisible by $3$.

Alternatively, each time Rule 2 is used, simply double all the coordinates before making the jump. This cannot change whether a frog can reach the origin, and it ensures that the coordinates remain integers and that the jumps are all by a multiple of $3$, so the same argument for Rule 1 still works.