#### Answer

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

#### Work Step by Step

(a) Vertex C and vertex I are odd vertices. All other 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 I. From there, the path must travel to vertex H, and then to vertex D.
At this point, we have a decision to make. According to Fleury's Algorithm, we should always choose an edge that is not a bridge if possible. Since the edge CD is a bridge, we should choose a different edge.
From vertex D, the path can travel to vertex F, then to vertex G, then to vertex E, and then back to vertex D. From there, the path can travel to vertex C, then to vertex A, then to vertex B, and finally back to vertex C.
This path is I,H,D,F,G,E,D,C,A,B,C. 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.