### Is this statement correct?

Unless I am mistaken, it appears that there does not exist any for which any of the above problems has a positive solution. For instance, let be the complete graph with vertex set , and define to be the edge-cut of consisting of all edges between and . Now define a weighting of by assigning each edge in weight 1 and every other edge weight 2. So, the subgraph (consisting of all edges of weight 1) is isomorphic to - and since has edge-disjoint spanning trees (by the Nash-Williams theorem, say), our procedure may well choose trees so that all of the edges in all of these graphs are in . But then will not have a Hamiltonian path since it is a subgraph of .

Of course, we may modify the edge weights here so that the procedure is forced to choose so that all of these trees have their edges in .