Graphs with 12 edges over the vertices $\{1,2,...,12\}$ have two vertices with a degree of 5 
How many graphs with 12 edges over the vertices $\{1,2,...,12\}$ have two vertices with a degree of 5?

The two vertices aren't neighbours: $\binom {10} 2 \binom 85 ^2 \binom {\binom 82} 2$. Explanation: choosing the two, then neighbours for each, then a place for the two edges that's left.
The two vertices are neighbours: $\binom {10} 2 \binom 8 4 ^2 \binom {\binom 82} 3$
In both cases there could be a third vertex with a degree 5 so we need to uncount it: $\binom {10} 3 \binom 7 3 ^3$, choosing the 3 and then since they're all neighbours to each other, choose another 3 neighbors for each, which is exactly 12 edges.
The total is:
$\binom {10} 2 \binom 85 ^2 \binom {\binom 82} 2 + \binom {10} 2 \binom 85 ^2 \binom {\binom 82} 2 - \binom {10} 3 \binom 7 3 ^3$
Does this cover all the overcounting? Is there a way to know for sure?
 A: Sorry to give a partial answer, but here is how I would find the cases for a subset of the (unlabelled) graphs.
Starting from the two situations you mention : the degree-5 vertices disconnected and connected to each other. In both cases, we can assume a degree sequence of [5, 5, 1, ... 1] with 10 1's. This might be clearer from a diagram:
The key should explain, but these are the 5 cases from graphs with 10 edges. Unfortunately, we also have to consider extension of the one with 9 edges where the C subgraph is disconnected.
A: I don't really get why you subtracted 2each times.
For example when you count the number of graph with 3 vertices of degree 5 what I get is: $${12\choose 3}{9\choose 3}^3$$
Choose the three vertices that have degree 5. You know that they are neighbour hence there is 3 edges left to choose among the 9 vertices left and this 3 times (once for each vertex).
In the same way for the number of graph with 2 vertices with degree 5 I get:
First case the two vertex are neighbour:
$${12\choose 1}{11 \choose 5}{5\choose 1}{10\choose 4}{E(10)\choose 3}$$
Choose the first vertex, then it's five neighbour, choose the second vertex among the neighbour, choose the 4 neighbour left for the second vertex. Last choose the 3 reaming edges among the other vertices. Here $E(10)=(9*10)/2$, is the number of edges in the complete graph with $10$ vertices.
Second two vertex are not neighbour:
$${12\choose 1}{11 \choose 5}{6\choose 1}{10\choose 5}{E(10)\choose 2}$$
Choose the first vertex, then it's five neighbour, choose the second vertex among the 6 vertices that are not the first vertex nor one of its neighbour, then choose the 5 neighbour for the second vertex. Last choose the 2 reaming edges among the other vertices. Again $E(10)=(9*10)/2$, is the number of edges in the complete graph with $10$ vertices.
So I get:
$$
{12\choose 1}{11 \choose 5}{5\choose 1}{10\choose 4}{E(10)\choose 3}+{12\choose 1}{11 \choose 5}{6\choose 1}{10\choose 5}{E(10)\choose 2}
$$
Which is equal to 90 billion and something (see here).
May be your solution was correct, but what I'm sure of it is that your explanation didn't convince me at all. As I said: Why ${10 \choose 2}$ and not ${12 \choose 2}$? Why ${{8\choose 2} \choose 2}$ correspond to "the place for the two edges that's left"?
