if G is an Eulerian graph (which means it has Eulerian circle), and G also has an Eulerian path which is not a circle
there's not such graph right?
for Eulerian circle all vertex degree must be an even number, and for Eulerian path all vertex degree except exactly two must be an even number. and no graph can be both...
if in a simple graph G, a certain path is in the same time both an Eulerian circle and an Hamilton circle. it means that G is a simple circle, G is a circle or G is a simple graph, but it doesn't have to be circle? I really don't know... I think it means G is a simple circle but I can't explain why.
If $G$ has a cycle which is both Eulerian and Hamiltonian, it is made of $n$ distinct edges and $n$ distinct vertices. To have $|V| = |E| = n$, a graph must either have multiple components, or be a tree plus one edge. Show that neither of those is going to work, except for a simple circle.