The main point : all your subgroups are rather small and
generated by a single rotation.
There are at least 24 elements in $G$ :
The identity ${\sf id}$,
Rotations around axes through the middles of one of
3 pairs of opposing faces and with a rotation of
$\frac{k\pi}{2}(1\leq k \leq 3)$ :there are $3\times 3=9$ such rotations,
and we denote them by $OF(<name\ of\ face>,<angle>)$
Half-turns around axes through the middles of one of
6 pairs of opposing edges : there are $6$ such rotations, and we denote them by $HT(<name\ of\ opposite\ edges>)$.
Rotations with axis $18,26,37$ or $45$, and angle
$\frac{2k\pi}{3}(1\leq k \leq 2)$ :there are $2\times 4=8$ such rotations,
and we denote them by $R(<name\ of\ axis>,<angle>)$.
Since there are at most $24$ orientation-preserving of the cube
(see here), we see that $G$ consists exactly of the $24$ elements
enumerated above.
Let $G_v$ denote the subgroup of $G$ fixing the vertex $1$, $G_e$ the subgroup
fixing the edge $12$, and $G_f$ the subgroup fixing the face $1234$.
By the enumeration above, we have :
$$
\begin{array}{lcl}
G_v &=& \lbrace {\sf id};R(18,\frac{2k\pi}{3}) (1 \leq k \leq 3) \rbrace \\
G_e &=& \lbrace {\sf id};HT(12,78) \rbrace \\
G_f &=& \lbrace {\sf id};OF(1234,\frac{k\pi}{2}) (1 \leq k \leq 3) \rbrace
\end{array}
$$
It is easy then to deduce the decompositions into orbits :
$$
\begin{array}{|l|l|}
\hline
\text{Subgroup} & G_v \\
\hline
\text{Orbits in } V &
[1] [2,3,5] [4,6,7] [8] \\
\hline
\text{Orbits in } E &
[12,13,15] [24,37,56] [26,34,57] [48,68,78] \\
\hline
\text{Orbits in } F &
[1234,1256,1357] [2468,3478,5678] \\
\hline
\end{array}
$$
$$
\begin{array}{|l|l|}
\hline
\text{Subgroup} & G_e \\
\hline
\text{Orbits in } V &
[1,2] [3,6] [4,5] [7,8] \\
\hline
\text{Orbits in } E &
[12] [13,26] [15,24] [34,56] [37,68] [48,57] [78] \\
\hline
\text{Orbits in } F &
[1234,1256] [1357,2468] [3478,5678] \\
\hline
\end{array}
$$
$$
\begin{array}{|l|l|}
\hline
\text{Subgroup} & G_f \\
\hline
\text{Orbits in } V &
[1,2,3,4] [5,6,7,8] \\
\hline
\text{Orbits in } E &
[12, 13, 24, 34] [15, 26, 37, 48] [56, 57, 68, 78] \\
\hline
\text{Orbits in } F &
[1234] [1256,1357,2468,3478] [5678] \\
\hline
\end{array}
$$
This one can be done with the Polya Enumeration Theorem, which requires the cycle index $Z(G)$ of the face permutation group $G$ of the cube. We now enumerate the permutations from this group by their cycle structure.
There is the identity, which contributes $$a_1^6.$$
There are rotations about diagonals connecting opposite vertices, which contribute
$$4 \times 2 \times a_3^2.$$
There are the rotations about an axis passing through the centers of opposite faces, which contribute $$3\times (2 a_1^2 a_4 + a_1^2 a_2^2).$$
Finally there are rotations about an axis passing through the midpoints of opposite edges, which contribute
$$6\times a_2^3.$$
It follows that the cycle index is
$$Z(G) = \frac{1}{24} a_1^6 + \frac{1}{3} a_3^2 +
\frac{1}{4} a_1^2 a_4 + \frac{1}{8} a_1^2 a_2^2 + \frac{1}{4} a_2^3.$$
Substituting the three colors red, green, and blue into this cycle index we get
$$Z(G)(R+G+B) =
1/24\, \left( R+G+B \right) ^{6}+1/4\, \left( R+G+B \right) ^{2} \left( {R}^{4}
+{G}^{4}+{B}^{4} \right)\\ +1/8\, \left( R+G+B \right) ^{2} \left( {R}^{2}+{G}^{2
}+{B}^{2} \right) ^{2}+1/3\, \left( {R}^{3}+{G}^{3}+{B}^{3} \right) ^{2}+1/4\,
\left( {R}^{2}+{G}^{2}+{B}^{2} \right) ^{3}.$$
Expanding this cycle index we obtain
$${B}^{6}+{B}^{5}G+{B}^{5}R+2\,{B}^{4}{G}^{2}+2\,{B}^{4}GR+2\,{B}^{4}{R}^{2}+2\,{
B}^{3}{G}^{3}+3\,{B}^{3}{G}^{2}R+3\,{B}^{3}G{R}^{2}\\+2\,{B}^{3}{R}^{3}+2\,{B}^{2
}{G}^{4}+3\,{B}^{2}{G}^{3}R+6\,{B}^{2}{G}^{2}{R}^{2}+3\,{B}^{2}G{R}^{3}+2\,{B}^
{2}{R}^{4}\\+B{G}^{5}+2\,B{G}^{4}R+3\,B{G}^{3}{R}^{2}+3\,B{G}^{2}{R}^{3}+2\,BG{R}
^{4}+B{R}^{5}+{G}^{6}+{G}^{5}R\\+2\,{G}^{4}{R}^{2}+2\,{G}^{3}{R}^{3}+2\,{G}^{2}{R
}^{4}+G{R}^{5}+{R}^{6},$$
which finally gives $$[R^2G^2B^2] Z(R+G+B) = 6.$$
This generating function includes all distributions of three or fewer colors,e.g. the coefficient of $GR^5$ is one because there is only one way to paint the cube using one color five times and a second one for the remaining face. Note that we had rotations about face pairs, edge pairs and vertex pairs, which is a standard feature in the symmetry groups of regular polyhedra. The reader is invited to compute the cycle index of the symmetry group of the faces of a tetrahedron.
Here is another interesting calculation involving a cycle index.
Best Answer
The point of the comment in the book is this: given a cube, pick one corner and call it $A$. You can certainly rotate the cube so that $A$ either stays put or moves to any chosen other corner, of which there are 7. So you have 8 choices for where you want $A$ to be.
Once you've done this, the other corners are obviously not free to move as they wish. Let $B$ be some corner adjacent to $A$, meaning it shares an edge of the cube with $A$. If A moves to $A'$, $B$ must move to some corner adjacent to $A'$ - otherwise the edge $AB$ gets distorted. How many corners are adjacent to $A'$? That's where the 3 comes from.
Finally you'll need to convince yourself that once $A$ and $B$ have been placed, all other corners are "forced" - their final location is already fixed. See if you can do that just using the relationships of adjacency, sharing a face etc.