Tính a + b .

Do \(\frac{\pi }{2} < \alpha ,\beta  < \pi  \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\cos \alpha  < 0}\\{\sin \beta  > 0}\end{array}} \right.\).

Ta có \(\cos \alpha  =  - \sqrt {1 - {{\sin }^2}\alpha }  =  - \sqrt {1 - \frac{1}{9}}  =  - \frac{{2\sqrt 2 }}{3};\sin \beta  = \sqrt {1 - {{\cos }^2}\beta }  = \sqrt {1 - \frac{4}{9}}  = \frac{{\sqrt 5 }}{3}\).

Suy ra \(\sin \left( {\alpha  + \beta } \right) = \sin \alpha  \cdot \cos \beta  + \cos \alpha  \cdot \sin \beta  = \frac{1}{3} \cdot \left( { - \frac{2}{3}} \right) + \left( { - \frac{{2\sqrt 2 }}{3}} \right) \cdot \frac{{\sqrt 5 }}{3} =  - \frac{{2 + 2\sqrt {10} }}{9}\).

Suy ra \(\sin \left( {\alpha  + \beta } \right) =  - \frac{{2 + 2\sqrt {10} }}{9} =  - \frac{{2\left( {1 + \sqrt {10} } \right)}}{9}\).

Vậy\(a + b = 2 + 9 = 11\).

Đáp án: 11.