jmc

Making all of the quantities have a base of 3 will simplify comparison. $$3^{-2}=3^{-2}$$ $$9^{-2}=(3^2{)}^{-2}=3^{2\cdot -2}=3^{-4}$$ $$27^{-\frac{2}{3}}=(3^3{)}^{-\frac{2}{3}}=3^{3\cdot -\frac{2}{3}}=3^{-2}$$ $$9\cdot 81^{-1}=3^2\cdot (3^4{)}^{-1}=3^2\cdot 3^{4\cdot -1}=3^{2+(-4)}=3^{-2}$$ $$243^{-\frac{4}{5}}=(3^5{)}^{-\frac{4}{5}}=3^{5\cdot -\frac{4}{5}}=3^{-4}$$ The two different values are $3^{-2}$ and $3^{-4}$, so our answer is $\boxed{2}$.