Let $T$ be a local minimum spanning tree that is not a global minimum.
Let $e_1,\ldots,e_m$ the edges of $T$ ordered by non-decreasing weight.
Let $T'$ be a minimum weight spanning tree that coincides with $T$ on the largest possible begin segment,
so we may assume $T'$ has edges $e_1,\ldots,e_{n-1},f_n,\ldots,f_m$ (again ordered by non-decreasing weight)
and $f_n\ne e_n$. Let $w$ be the weight function in our graph.
We claim that $w(f_n)<w(e_n)$: indeed, if $w(e_n)\leq w(f_n)$ then $e_1,\ldots,e_{n-1},e_n$ would
be a valid startup for Kruskal's algorithm and would lead to a minimum weight spanning tree that
coincides with $T$ on a larger initial segment.
$T+f_n$ contains exactly one cycle $C$. The edges of $C$ different from $f_n$ cannot all be in
$\{e_1,\ldots,e_{n-1}\}$ or we would have a cycle in $T'$. So we find at least one edge $f$ on $C$ that
is different from $e_1,\ldots,e_{n-1}$ and $f_n$. But then $w(f)\geq w(e_n)>w(f_n)$,
so $T+f_n-f$ is a spanning tree with a smaller total weight that is a neighbour of $T$ in the spanning tree graph. Contradiction.
(I left some small details for you to prove. Let me know if they cause trouble).
Note that this proves the slightly stronger statement, that each non-minimal tree has
a neighbour of strictly smaller total weight.
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
I will prove the following Lemma, which answers the question you are asking.
To prove this, let $T_1'$ be any other spanning tree of $G[W]$. Consider the graph $T'$, with vertex set $V$, whose edge set $E(T')$ is given by $$ E(T')=(E(T)\setminus E(T_1))\cup E(T_1') $$
That is, take $T$, remove the edges of $T_1$, then add in the edges of $T_1'$. I claim that $T'$ is a spanning tree of $G$. To see this, note
$T'$ is connected. Indeed, for any vertices $v_1,v_2\in V$, there is a path in $T$ connecting them, which can be modified to be a path in $T'$ by replacing the portion of the path that lies in $T_1$ with a corresponding path in $T_1'$.
$T'$ has the same number of edges as $T$.
Since $T'$ is a connected graph with $|V|-1$ edges, it is a tree on the same vertex set as $G$, and therefore a spanning tree.
Now, since $T$ was an MST for $G$, we conclude that $w(T')\ge w(T)$. Since $w(T)=w(T_1)+w(T\setminus T_1)$, and $w(T')=w(T_1')+w(T\setminus T_1)$, it follows that $w(T_1')\ge w(T_1)$, thus proving the minimality of $T_1$.