Воспользуемся основным тригонометрическим тождеством:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {-\dfrac{{20}}{{29}}} \right)^2} + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\cos ^2}\alpha = \dfrac{{441}}{{841}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\cos \alpha = \dfrac{{21}}{{29}},\,\,\,}\\{\cos \alpha = -\dfrac{{21}}{{29}}.}\end{array}} \right.\)
Так как \(\pi < \alpha < \dfrac{{3\pi }}{2}\) (III четверть), то \(\cos \alpha < 0\), то есть \(\cos \alpha = -\dfrac{{21}}{{29}}.\)
\(29\sqrt 8 \sin \left( {\alpha + \dfrac{\pi }{4}} \right) = 29\sqrt 8 \left( {\sin \alpha \cos \dfrac{\pi }{4} + \cos \alpha \sin \dfrac{\pi }{4}} \right) = \)
\( = 29\sqrt 8 \left( {-\dfrac{{20}}{{29}} \cdot \dfrac{{\sqrt 2 }}{2}-\dfrac{{21}}{{29}} \cdot \dfrac{{\sqrt 2 }}{2}} \right) = -\dfrac{{29\sqrt 8 \cdot \sqrt 2 }}{2} \cdot \dfrac{{41}}{{29}} = -82.\)
Ответ: \(-82.\)