Воспользуемся основным тригонометрическим тождеством:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha + {\left( {\dfrac{3}{5}} \right)^2} = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha = \dfrac{{16}}{{25}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \alpha = \dfrac{4}{5},}\\{\sin = -\dfrac{4}{5}.}\end{array}} \right.\)
Так как \(\dfrac{{3\pi }}{2} < \alpha < 2\pi \) (IV четверть), то \(\sin \alpha < 0\), то есть
\({\sin ^2}\beta + {\cos ^2}\beta = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\beta + {\left( {-\dfrac{4}{5}} \right)^2} = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\beta = \dfrac{9}{{25}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \beta = \dfrac{3}{5},\,\,\,}\\{\sin \beta = -\dfrac{3}{5}.}\end{array}} \right.\)
Так как \(\pi < \beta < \dfrac{{3\pi }}{2}\) (III четверть), то \(\sin \beta < 0\), то есть \(\sin \beta = -\dfrac{3}{5}.\)
\(\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta -\sin \alpha \sin \beta = \dfrac{3}{5} \cdot \left( {-\dfrac{4}{5}} \right)-\left( {-\dfrac{4}{5}} \right) \cdot \left( {-\dfrac{3}{5}} \right) = -\dfrac{{12}}{{25}}-\dfrac{{12}}{{25}} = -\dfrac{{24}}{{25}} = -0,96.\)
Ответ: \(-0,96.\)