1. It should be correct as it is present in all the option.
2. For cycle graph (2e/n) = (2n/n) = 2. It is true because remove the opposite vertex, graph will become disconnected.
For complete graph (2e/n) = (n(n-1)/n) = (n-1). It is true.
So this option seem to be correct.
3. ORE's Theorem: If G is a simple graph with n vertices with n>= 3 such that deg(u) + deg(v) >= n for every pair of non-adjacent vertices u and v in G, then G has a hamiltonian cycle.
3 is wrong. Because It is sufficient condition not necessary. A graph having this deg(u) + deg(v) >= n as true will definitely have hamiltonian circuit. But it doesn't mean that a graph have not this condition deg(u) + deg(v) >= n as true will not have a hamiltonian circuit. Ex. A cycle graph with n = 6 will have hamiltonian circuit.
4. Edge-disjoint hamiltonian circuit : Hamiltonian circuit which doesn't have any common edge with other hamiltonian circuit in same graph.
In complete graph for n = 3. We have only one hamiltonian circuit. (3-1)/2 = 1.
For n = 4. We have 6 edges in complete graph. Suppose in one hamiltonian cycle you include e1, e2, e3, e4 edges. Now we have e5 and e6 edge left we can't for another hamiltonian cycle with this edge. So again one.
(4 - 1)/ 2 = 1.
So Answer: C.