Complexity of counting weighted cycles in planar graphs?
If you allow repetition of vertices, but not edges, within a cycle, then the problem is $\text{#P}$-complete.  This paper shows that the problem of counting Eulerian tours is $\text{#P}$-complete for 4-regular planar graphs.  Therefore the problem I just described (call it $\text{#}A^\prime$) is $\text{#P}$-hard, because it is at least that difficult: in the special case where all edges have weight 1 and all nodes have weight 0, the Eulerian tours are just those cycles with total weight equal to the number of edges in the graph.  On the other hand, $\text{#}A^\prime\in\text{#P}$, because the corresponding decision problem $A^\prime \in \text{NP}$.