jmc

Let $m=10!+2$ and $n=11!+8$. Then, $n-11m=(11!+8)-11(10!+2)=8-22=-14$. By the Euclidean algorithm, $$\text{gcd}(m,n)=\text{gcd}(m,n-11m)=\text{gcd}(m,-14).$$Since $7$ divides $10!$, it follows that $7$ does not divide $10!+2$. However, $10!+2$ is even, so it follows that $\text{gcd}(m,14)=\boxed{2}$.