#### Answer

(a) We need to add the edges AB, AC, BC, and DE to make this graph a complete graph. There are 24 Hamilton circuits in this complete graph.
(b) The paths A,B,C,D,E,A and A,E,C,D,B,A are Hamilton circuits in the modified graph.
(c) If we remove the edges AD and AE, then all the vertices will be even. Then the graph will have at least one Euler circuit.
(d) The path B,D,C,E,B is an Euler circuit in the modified graph.

#### Work Step by Step

(a) In a complete graph, there is an edge between every pair of vertices in the graph. We need to add the edges AB, AC, BC, and DE to make this graph a complete graph.
The number of Hamilton circuits in a graph with $n$ vertices is $(n-1)!$. The number of Hamilton circuits in this graph is $(5-1)! = 4! = 24$. There are 24 Hamilton circuits in this complete graph.
(b) A Hamilton circuit is a path that passes through each vertex exactly once, and it starts and ends on the same vertex. The paths A,B,C,D,E,A and A,E,C,D,B,A are Hamilton circuits in the modified graph.
(c) For a graph to have an Euler circuit, the number of odd vertices must be 0. In the given graph, vertex D and vertex E are odd vertices. If we remove the edges AD and AE, then all the vertices will be even. Then the graph will have at least one Euler circuit.
(d) An Euler circuit is a path that travels through every edge in the graph exactly once, and the path starts and ends at the same vertex. The path B,D,C,E,B is an Euler circuit in the modified graph.