First thing to immediately take notice here, is that edges have distinct weights. This characteristic changes results.
Whenever it isn't mentioned that edge weights are distinct, you always should keep all edges at same weight to fetch an extreme case, which will generate counter-examples in a lot of graph problems.
Now, coming to the question:
If e is the lightest edge of some cycle in G, then every MST of G includes e.
The only reason a lighter edge is not preferred in some MST is when it somehow forms a loop. So, try to make a loop out of it.
If this lightest edge of some cycle contributes to a loop somewhere else, then we will exclude it.
See this:
50 is the lightest edge of some cycle. But we will exclude it because it'll form a loop in our MST.
Hence, Statement I is False
If e is the heaviest edge of some cycle in G, then every MST of G excludes e.
The edge weights are distinct. It means there's only one heaviest edge strictly.
The only reason a heavier edge is included in an MST is when it is a bridge (cut edge). Including it is our necessity because without it, the MST can't be connected.
Also, it is a property that a bridge can't be a part of a cycle.
This means given heaviest edge is NOT a bridge. So including it is not our necessity. And when it's not necessary, we certainly won't include it in our MST — we'll always prefer a lighter edge.
Statement II is True.
Option B
Please note that this statement is true only because the edge weights are distinct.
If not, then there exists a graph where many, if not all edge weights are the heaviest — in such a case we will be including multiple max-weight edges, lol. (Take a graph with one edge of weight 1, and all other edges of weight 500)
https://gateoverflow.in/3813/gate2005-it-52