So, basically neither BFS or DFS yield cross edges. So (I) is wrong anyways.
In (II) condition we can have edges in the graph which are at the same level in the corresponding BFS traversal.
Say, node 1 is parent and node 2 and 3 are its children. So in DFS we will have edges (1,2) and (1,3). Here node 2 and node 3 are at the same depth i.e. depth 1, they are 1 level below node 0, while node 0 is at 0 depth (it is start of traversal). thus depth(1)-depth(2)=1-1=0. Statement (II) says it can only be 1. Thus it is false.