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.
I'll first consider the case with no parallel edges. Considering Kruskal's algorithm edges with weights 5 and 8 would be taken with any possible structure of the graph. Then if edges 5, 8 and 10 doesn't form a cycle Kruskal's algorithm will take edge with weight 10 and finish its work. Hence in optimal graph edges 5, 8 and 10 form a cycle. I'll denote the vertices of this cycle as $A$, $B$ and $C$. The fourth vertex will be denoted as $D$ Then edge of weight 16 have to connect $D$ with $A$ or $B$ or $C$ since parallel edges are not allowed. So, Kruskal's algorithm will take it and finish its work. The answer in that case is $5+8+16=29$
The case with parallel edges is more obvious. Kruskal's algirithm have to take edge with weight 5. Then graph with vertices $\{A,B,C,D\}$ and three edges between $A$ and $B$ with weights 5,8 and 10, edge between $B$ and $C$ with weight 16 and edge between $C$ and $D$ with weight 18 gives us maximal possible answer.
Best Answer
Hint: The algorithm goes this way: Choose the edges weight from the lowest to highest. That edge will be added if it doesnt form a cycle with already choosen edges. The algorithm stops when a spanning tree is formed.