jmc

If we first get a "common denominator" as if 2, 5, 8, and 11 represent real numbers rather than residues, we get $$\frac{2}{5}+\frac{8}{11}\equiv \frac{2\cdot 11+8\cdot 5}{55}\equiv \frac{62}{-1}\equiv -62\equiv \boxed{50}(\mathrm{mod}56).$$Indeed, we can justify this manipulation as follows. Suppose that $n\equiv 2\cdot 5^{-1}+8\cdot 11^{-1}(\mathrm{mod}56)$; then multiplying both sides of the congruence by $55$ (which is relatively prime with $56$) yields that $-n\equiv 55n\equiv 22+40\equiv 62(\mathrm{mod}56)$.