How would I go about solving this?
I know that $K_{10}$ has $9+8+7+\dots+1=45$ edges.
So would it be something like $\binom {45}{43}$ because out of the 45 total edges, one must choose 43 for the graph?
Second attempt: At least for the second bullet point mentioned, it seems that there would be $\binom{10}{2}9\cdot7$ ways. There are two vertices to choose from ten. Then, for the first vertex, there are nine options for a connecting vertex, and for the second vertex, there are 7 options for a connecting vertex.}
Third attempt: I don't know what I was thinking ^. I can only find one non-isomorphic class for the second bullet point, and they are not isomorphic those in the first class.
HINTS: The question is: how many non-isomorphic graphs can you get by removing two edges from $K_{10}$?
You can remove two edges that share a vertex. There are $10\binom92=360$ ways to do this, so you get $360$ graphs, but you should be able to show that they are all isomorphic.
You can remove two edges that do not share a vertex. Are all of the resulting graphs isomorphic to one another? Are they isomorphic to those of the first type?
Added: As DKal suggested in the comments, it may be helpful to look at the complements of the resulting graphs, if you already know (or can figure out how to prove) that two graphs are isomorphic if and only if their complements are isomorphic.