Воспользуемся основным тригонометрическим тождеством:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {\dfrac{5}{{13}}} \right)^2} + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\cos ^2}\alpha = \dfrac{{144}}{{169}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\cos \alpha = \dfrac{{12}}{{13}},\,\,\,}\\{\cos \alpha = -\dfrac{{12}}{{13}}.}\end{array}} \right.\)
Так как \(\dfrac{\pi }{2} < \alpha < \pi \) (II четверть), то \(\cos \alpha < 0\), то есть \(\cos \alpha = -\dfrac{{12}}{{13}}.\)
\({\sin ^2}\beta + {\cos ^2}\beta = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\beta + {\left( {-\dfrac{5}{{13}}} \right)^2} = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\beta = \dfrac{{144}}{{169}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \beta = \dfrac{{12}}{{13}},\,\,\,}\\{\sin \beta = -\dfrac{{12}}{{13}}.}\end{array}} \right.\)
Так как \(\pi < \beta < \dfrac{{3\pi }}{2}\) (III четверть), то \(\sin \beta < 0\), то есть \(\sin \beta = -\dfrac{{12}}{{13}}.\)
\(169\sin \left( {\alpha + \beta } \right) = 169\left( {\sin \alpha \cos \beta + \cos \alpha \sin \beta } \right) = \)
\( = 169\left( {\dfrac{5}{{13}} \cdot \left( {-\dfrac{5}{{13}}} \right)-\dfrac{{12}}{{13}} \cdot \left( {-\dfrac{{12}}{{13}}} \right)} \right) = 169 \cdot \left( {\dfrac{{-25}}{{169}} + \dfrac{{144}}{{169}}} \right) = 169 \cdot \dfrac{{119}}{{169}} = 119.\)
Ответ: 119.