How to find the number of different m-colorings of vertices? 
These are the textbook answers for d) and e), but I'm not exactly sure how they get it. Can someone show me with diagram like a picture perhaps. Thank you. You can just show me one part, don't need to be both parts.

We will do the second problem. This question looks like it is asking for
the Burnside lemma. The term  floating suggests rotational symmetries
only but there are twelve terms in the proposed closed form which means
we probably have dihedral  symmetry. Recall  the cycle index of the
cyclic group:
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$
as well as the cycle index of the dihedral group
$$Z(D_n) =
\frac{1}{2} Z(C_n) +
\begin{cases}
\frac{1}{2} a_1 a_2^{(n-1)/2} & n \text{ odd} \\
\frac{1}{4} \left( a_1^2 a_2^{n/2-1} + a_2^{n/2} \right)
& n \text{ even.}
\end{cases}$$
Therefore we get for the cycle index of the hexagon without the central
node
$$Z(D_6) = \frac{1}{12} (a_1^6 + a_2^3 + 2 a_3^2 + 2 a_6)
+ \frac{1}{4} (a_1^2 a_2^2 + a_2^3).$$
The central node is always fixed so we get for our cycle index of the
wheel
$$Z(W_6) = \frac{a_1}{12} (a_1^6 + a_2^3 + 2 a_3^2 + 2 a_6)
+ \frac{a_1}{4} (a_1^2 a_2^2 + a_2^3).$$
By Burnside we must be constant on the cycles and hence we have $m$
choices for every cycle:
$$\frac{m}{12} (m^6 + m^3 + 2 m^2 + 2 m)
+ \frac{m}{4} (m^4 + m^3)
\\ = \frac{1}{12} (m^7 + m^4 + 2 m^3 + 2 m^2 + 3 m^5 + 3 m^4)
\\ = \frac{1}{12} (m^7 + 3 m^5 + 4 m^4 + 2 m^3 + 2 m^2).$$
This is the claim.
