Take any spanning tree T of G. For any non-tree edge e, adding e to T creates a unique cycle. If we remove any tree edge f on that cycle, we get another spanning tree.
Since every spanning tree has even total weight,
w(T)≡0(mod2)andw(T)−w(f)+w(e)≡0(mod2).
Hence,
w(e)≡w(f)(mod2).
So, within any cycle, all edges must have the same parity. They may all be even or all be odd. Therefore (D) is true.
The other statements are not forced in general.
Therefore, the correct answer is D.