So the concatenation of these three bijections between E(G) and E(H) maps each cycle to a cycle and each cut to a cut and vica versa. Therefore G and H are weakly isomorphic.
Their duals are isomorphic:
If the vertex numbers of the two trees are diffent, then their edge numbers are also different, therefore there cannot be a bijection between their edge set.
If their vertex numbers are the same, lets say n, then each of them, have n-1 edges.
A tree does not have a cycle and each edge of a tree is a cut. Therefore in this case any bijection between the edge sets maps a cut to a cut and a cycle to a cycle and vica versa. Therefore in this case they are weakly isomorphic.
By the 4 color theorem G can be colored by 4 colors. Color G with 4 colors.
Lets say the colors are 1,2,3 and 4. Each edge is incident to two different colors.
We divide the edge set to two sets:
E1: edges whose endpoints have colors (1,2), (2,3), (1,4) E2: edges whose endpoints have colors (1,3), (3,4), (2,4)
(V,E1) is bipartite: To show this we show that it does not contain an odd cycle.
Consider any cycle of this graph. If it does not contain a vertex colored by 1, then all of its vertices are colored by two colors so it is an even cycle. Otherwise start from a vertex colored by 1. We cannot move back to this color class by odd number of
edges, because after an edge colored (1,2) we need even number of edges colored (2,3) to be at a vertex colored by 2 and then we need an edge colored (1,2) to arrive back to a vertex colored 1. This is an even number. If we have an edge (1,4), then by the next edge we are back at 1. So between any two vertices of the cycle which are colored 1 there are even number of edges, in both directon, so the cycle is even.
(V,E2) is bipartite. The proof is similar. (If it does not contain 4 as a colro, then it uses two colors. etc.)
G* is connected. Since G is the dual of G*, G is connected. Therefore they do not
have isolated vertices. G does not have a leaf, because a leaf would imply a loop in G*, but G* is simple. G does not have a degree two vertex, because if G has a degree
two vertex, then the incident edges are serial edges, therefore in G* the corresponding edges would be paralell edges, but G* is simple.
Similarly G* does not have a degree one or two vertex.
What if both minimum degree is at least 4? Indirectly assume this.
This would mean that each face of G and G* have at least 4 edges. Therefore neither G nor G* contain a cycle of length 3 or less. They are planar, so:
Let F1 be the number of edges bound the first face, F2 be the number of edges bound the second face, etc.
Lets count the edges:
Therefore either G or G* contains a degree 3 vertex and we are done.
If G has 3n-6 edges and planar, then each face of G is a triangle. Proof: If it has a face which is not a triangle, then adding a diagonal of that increasing the edge number but keeping the graph planar. But a graph having 3n-5 edges is not planar.
So the degree of each vertex of G* is 3, therefore its maximum degree is 3.
