As you correctly state, the graph has not Hamiltonian Cycle, since it has a cut vertex (actually it has two, namely $c$ and $f$).
Regarding the number of distinct Eulerian Circuits, there are more than two if you take the definition of distinct to be
Two Eulerian Circuits are distinct, iff they are not exact reversals of each other.
In this case, the three circuits
$$(a,c,d,f,g,h,f,e,c,b,a)$$
$$(a,c,d,f,h,g,f,e,c,b,a)$$
$$(a,c,e,f,g,h,f,d,c,b,a)$$
are examples of distinct Eulerian Circuits.
As stated in Brian's answer, there are exactly 4 distinct Eulerian Cycles, namely
$$a,c,d,f,g,h,f,e,c,b$$
$$a,c,d,f,h,g,f,e,c,b$$
$$a,c,e,f,g,h,f,d,c,b$$
$$a,c,e,f,h,g,f,d,c,b$$
To count these, and make sure there are no more than 4, look at the number of times the walk "crosses itself". Since there are only two vertices ($c$ and $f$) where such a crossing can happen, the number of crossings can be 0,1 or 2. There is only a single walk with no crossing, namely
$$a,c,d,f,g,h,f,e,c,b$$
There is also only a single walk with two crossings, namely
$$a,c,e,f,g,h,f,d,c,b$$
and there are two walks with a single crossing, the one with a crossing in $c$
$$a,c,e,f,h,g,f,d,c,b$$
and the one with a crossing in $f$
$$a,c,d,f,h,g,f,e,c,b.$$