#### Answer

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

#### Work Step by Step

(a) Vertex F and vertex G are odd vertices. The other vertices are even vertices. This 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 F. According to Fleury's Algorithm, we should always choose an edge that is not a bridge, if possible. Since the edges FB, FD, and FG are not bridges, we can choose either of these edges as the next step in the path.
The path can travel to vertex G, then to vertex C, then to vertex B, and then back to vertex F.
From there, the path must travel to vertex D, then to vertex B, then to vertex A, then to vertex C, then to vertex E, and then finally back to vertex G, because these are the only available edges.
This path is F,G,C,B,F,D,B,A,C,E,G. 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.