jmc

First, let's ignore the requirement that the product can't be $6$. Then I can pick the first blue face in $6$ ways, and the second blue face in $5$ ways, making $6\cdot 5=30$ choices in all. But we've actually counted each possible result twice, because it makes no difference which of the two blue faces I chose first and which I chose second. So the number of different pairs of faces is really $(6\cdot 5)/2$, or $15$.

Now we exclude the pairs which have a product of $6$. There are two such pairs: $\{1,6\}$ and $\{2,3\}$. That leaves me $\boxed{13}$ pairs of faces I can paint blue.