#### Answer

(a) This connected graph has exactly two odd vertices. Therefore, by Euler's theorem, the graph has at least one Euler path.
(b) E,B,A,E,C,B,D,C,A,D is an Euler path.

#### Work Step by Step

(a) Vertex D and vertex E are odd vertices. The other three vertices in the graph are even vertices. This connected graph has exactly two odd vertices. Therefore, by Euler's theorem, the graph has at least one Euler path.
(b) If a graph has exactly two odd vertices, then any Euler path starts at one odd vertex and ends at the other odd vertex.
Let's start at vertex E. From there, the path can travel to vertex B, then to vertex A, and then to back to vertex E.
From there, the path can travel to vertex C, then to vertex B, and then to vertex D. At this point, there are three edges in the graph which have not been used yet. From vertex D, the path can travel to vertex C, then to vertex A, and finally back to vertex D.
This path is E,B,A,E,C,B,D,C,A,D. This path travels through every edge of the graph exactly once, so it is an Euler path.
This is one Euler path but there are other Euler paths in this graph also.