I suppose it's a little bit late to help you for your test, but here goes.
First, when you write "minimum spanning tree," all spanning trees in a given graph have the same number of nodes and the same number of edges. So I'm going to assume that what we are dealing with is a weighted graph, and we are discussing minimal weight spanning trees.
Now $(V',E'\cap T)$ clearly has no cycles, so it can be extended to a spanning tree of $G'$, perhaps in many ways. Among those ways, pick one, call it $T^+$ of minimal weight: we want to show $T^+$ is a minimum weight spanning tree for $G'$.
Well, suppose it isn't. Then there's a different tree, $T^-$, that is a spanning tree for $G'$ and has smaller weight than $T^+$. Now prove that you can extend $T^-$ to be a spanning tree $T^*$ for $G$ by adding some of the edges of $T$ (this, I think, is where you're going to have to add some details to fill out this sketch). But then $T^*$ has weight less than that of $T$, contradiction.
For (b), since we have proved $(V',E'\cap T)$ is a subgraph of a minimum spanning tree of $G'$, it follows that it is a minimum spanning tree of $G'$ if and only if it is a spanning tree of $G'$.
So suppose I have three disjoint sets of vertices: $\{v_{1}\} \cup \{v_{2}\} \cup V(C_{3})$. Here, $\{v_{1} \} \cup \{v_{2}\}$ is a forest which does not span, while $\{v_{1}\} \cup \{v_{2}\} \cup (C_{3} - e)$ is a spanning forest, for $e \in E(C_{3})$.
Best Answer
"Spanning" is the difference: a spanning subgraph is a subgraph which has the same vertex set as the original graph. A spanning tree is a tree (as per the definition in the question) that is spanning.
For example:
has the spanning tree
whereas the subgraph
is not a spanning tree (it's a tree, but it's not spanning). The subgraph
is also not a spanning tree (it's spanning, but it's not a tree).