An Euler circuit by definition traverses every edge of the graph exactly once, so neither $\langle a,b,c\rangle$ nor $\langle d,e,f\rangle$ is an Euler circuit. Starting at $a$ and counting two circuits as the same if one is simply the reversal of the other, I get the following four distinct Euler circuits:
$$\begin{align*}
&a,c,d,f,g,h,f,e,c,b\\
&a,c,d,f,h,g,f,e,c,b\\
&a,c,e,f,g,h,f,d,c,b\\
&a,c,e,f,h,g,f,d,c,b
\end{align*}$$
You are correct about the lack of a Hamilton circuit in this graph and the reason for it.
Here we go:
$$\huge\cdot\qquad\cdot$$
remember that $0$ is even. The circuit is the "empty circuit"
Since the graph has no edges, we've already passed every edge if we don't even move :D
Best Answer
You are correct.
Notice that such an Euler circuit must use $e_2$. So you can reduce the problem to finding the number of Euler trails starting at $C$ and ending at $A$ in $G - D$.
Think of an Euler trail in this case as a sequence of five edges $e_{i_1}e_{i_2}\cdots e_{i_5}$. Perhaps one of the easier ways to count is to single out an edge, say $e_3$, and count the number of valid sequences with $e_3$ in the first position, then the second, and so on. Since we are starting at $C$, you may notice that a sequence representing an Euler trail can only have $e_3$ in the first, third, and fifth position. You obtain
First: $4$ trails. Traverse $e_3$. There are $4$ ways to go from $A$ to $C$, back to $A$, that is two choices from $A$ to $B$, two choices from $B$ to $C$, and the way back is determined.
Third: $8$ trails. You can go $CBCABA$ of which there are four ways, or $CBACBA$, another four ways.
Fifth: $4$ trails, similar to the first case.
While this answer is long winded, at least in thinking this way you can be confident that you've exhausted all possibilities in your count. I believe in general it is hard ($NP$-hard?) to count the number of Euler circuits in a given undirected graph.