Technically, $C_n$ will have $n$ spanning trees ($n$ choices for the edge you you delete). But they are all isomorphic (paths of length $n-1$), so it depends on whether you want to consider them as distinct or the same.
Part $(b)$ is only true if we assume that $G$ has multiple edges but no loops.
The Wikipedia page is a little misleading; it might be more accurate to say that at least one of the lowest weight edges will be in a minimum spanning tree (in a loopless graph).
This is apparent from the application of Kruskal's Algorithm for constructing a minimum spanning tree. Essentially, select the lowest weight vertex at each step that will not introduce a cycle. Because it is impossible to create a cycle when selecting the first edge in a loopless graph (per assumption) Kruskal's Algorithm will always select it.
In order to rule out the length $10$,$10$, $11$, and $13$ edges, they must all create cycles if they are included with the $8$ edge; this implies that there is a multiple edge with $8$,$10$,$10$,$11$, and $13$ all connecting the same two vertices.
I would say it's very non-standard to allow multiple-edges but not loops in the graph, because problems are generally a matter of simple vs. non-simple, where loops are allowed in non-simple graphs. Clearly, you can obtain a higher result by making $8$,$10$,$10$, and $11$ edges all loops, while using the other $4$ to create the spanning tree.
I would ask your professor to clarify this point, and definitely ask during an exam if a question is vague.
Best Answer
Proof by contradiction:
Assume that adding an edge to a spanning tree does not create a cycle. Suppose we add edge $(u,v)$ between two vertices $u$ and $v$ to $T$, a spanning tree of a graph $G$. Since we assumed adding an edge to $T$ does not create a cycle, this implies that prior to the insertion of $(u,v)$, there was no path from $u$ to $v$ in $T$. But $T$ is a spanning tree, so such a path must have existed and we have arrived at a contradiction.
Therefore, adding an edge to a spanning tree creates a cycle.