Connections and Differences between these Cayley Diagrams for $A_4$ and $S_4$ - Carter pp. 80, 82 Reference: Nathan Carter pp. 80, 82, ch. 5, Visual Group Theory

Figure 5.24. As you will read in the next section, it is no
coincidence that [the Cayley digram for $S_4$] looks cube-like. A Cayley diagram for $S_4$arranged on a cube with clipped corners (usually called a "truncated cube"). The legend on the right shows which permutations in $S_4$ the arrows represent. Node labels were omitted to eliminate clutter.
Figure 5.27. A Cayley diagram for $A_4$ arranged on a truncated tetrahedron.
Figure 5.28. A Cayley diagram for $S_4$ arranged on a truncated octahedron.
The symmetry group for the tetrahedron is $A_4$, half of all permutations of four items.
We can lay out a Cayley diagram for $A_4$ in a way that makes its connection to the
tetrahedron clear; see Figure 5.27. The cube and octahedron have the same symmetry
group, S4. This is why it is no coincidence that the Cayley diagram for $S_4$ in Figure 5.24 looked cube-like.
(1.) There are no node labels hence I can't understand these Cayley diagrams?
And I don't understand how to fill them in?
(2.) I think I'm missing the hinges here. What are the important connections or differences between these diagrams? I don't see why "it is no coincidence that the Cayley diagram for $S_4$ in Figure 5.24 looked cube-like."
Please keep answers intuitive. I haven't covered orbits, cycles, alternating groups, cosets formally. 
 A: (1.) Arbitrarily pick a node and call it the identity, $e$. Starting from $e$, we can reach a node by following a red arrow. Call this node $r$, for example. Now starting again at $e$, we can also reach a node by following a blue line. Call this node $b$, for example. We can now label every node in terms of $r$ and $b$. For example, the triangle of red arrows that contains the identity refers to the nodes $e, r, r^2$. By examining one octagonal "face" that contains the identity (alternating between first red arrows then blue lines), we can make a round trip by traversing:
$$e,r,rb, rbr, rbrb, rbrbr, rbrbrb, rbrbrbr$$
Notice that $rbrbrbrb = e$.
Likewise, for the only other octagonal "face" that contains the identity (alternating between first blue lines then red arrows), we can make a round trip by traversing:
$$e,b,br, brb, brbr, brbrb, brbrbr, brbrbrb$$
Notice that $brbrbrbr = e$.

(2.) In the next section they discuss the platonic solids. To see why the cube and octahedron have the same symmetry group $S_4$, imagine taking each face $f$ of the cube and changing it into a vertex $\phi(f)$, then joining any two vertices $\phi(f)$ and $\phi(g)$ with an edge iff their original faces $f$ and $g$ share a common edge in the original cube. Then we obtain the octahedron (see the image at Wikipedia). This illustrates how the cube and the octahedron are said to be duals of each other.
