Given a spanning tree, cycle and edges, show a subgraph is a spanning tree 
Let $T, C$ be a spanning tree and  cycle in some graph $G$, respectively. Suppose $AB$ is an edge of $C$ and $T.$ Show that there is at least one other edge $UV \in C$ that can replace $AB \in T$ so that the resulting subgraph $T - AB + UV$ is still a spanning tree.

To start $T - AB$ is a two-component forest. If $C$ has no edge that can connect these components into a spanning tree in $G$, then $C$ must be disconnected in $G.$ Thus $T$ is not a spanning tree.
Not sure if it works. Would be happy if someone commented on it. Thanks.
 A: Not really sure where you're making the leap that $C$ must be disconnected in $G$, or why it's necessary.
The result pretty much follows by definition (if you're using the fairly standard Wilson text). A subgraph $T$ of a connected graph $G$ is a spanning tree if and only if each circuit of $G$ has an edge removed such that the graph stays connected.
So if $T$ is a spanning tree, then the circuit $C$ must have an edge $UV$ removed by hypothesis. Replacing $AB$ with $UV$ gives us a new spanning tree $T-AB+UV$.
EDIT 1: Turns out I was wrong in my initial conjecture that any such $UV$ would work. We need to be more careful. Here's what we need to do:
First, you want to verify your hint (if you want/need to) that $T-AB$ has two components $T_A$ and $T_B$ containing $A$ and $B$, respectively. This can easily be done by repeated use of Wilson Theorem 9A which states that a tree is equivalent with the property that "any two vertices of $T$ are connected by exactly one path".
Next, we identify the specific edge $UV \in C-T$ that we want, which turns out to be some $UV \in T$ such that $U$ and $V$ are in different components (we can then assume without loss of generality that $U \in T_A$ and $V \in T_B$). It is easy to figure why $T-AB+UV$ works once you build a new circuit $D$ (that isn't necessarily the same as $C$) in $G$ containing $AB$ and $UV$ that links $A$ to $U$ in $T_A$ and $B$ to $V$ in $T_B$. 
So we would really want such a $UV \notin T$ to exist. Denote the edges in $C$ as $C_0C_1, C_1C_2, \dots, C_{n-1}C_n$ for $n \geq 3$, where $C_{n-1}=A$ and $C_0 :=C_n :=B$. Note that $C_n$ is not in $T_A$ by definition. Let $m=min\{ 1 \leq i \leq n \colon C_i \notin T_A \}$. We find $C_{m-1}C_m$ necessarily follows our needed criteria.
EDIT 2: D41 follows directly from the last problem (since some $UV$ that follows our criteria exists for each $AB$ and $UV \notin T$ is hypothesized to be unique).
