Воспользуемся основным тригонометрическим тождеством:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha + {\left( {\dfrac{4}{{\sqrt {19} }}} \right)^2} = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha = \dfrac{3}{{19}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \alpha = \dfrac{{\sqrt 3 }}{{\sqrt {19} }},\,\,}\\{\sin \alpha = — \dfrac{{\sqrt 3 }}{{\sqrt {19} }}.}\end{array}} \right.\)
Так как \(\dfrac{{3\pi }}{2} < \alpha < 2\pi \) (IV четверть), то \(\sin \alpha < 0\), то есть \(\sin \alpha = -\dfrac{{\sqrt 3 }}{{\sqrt {19} }}.\)
\(\sqrt {19} \sin \left( {\alpha -\dfrac{\pi }{6}} \right) = \sqrt {19} \left( {\sin \alpha \cos \dfrac{\pi }{6}-\cos \alpha \sin \dfrac{\pi }{6}} \right) = \sqrt {19} \left( {-\dfrac{{\sqrt 3 }}{{\sqrt {19} }} \cdot \dfrac{{\sqrt 3 }}{2}-\dfrac{4}{{\sqrt {19} }} \cdot \dfrac{1}{2}} \right) = -\dfrac{3}{2}-\dfrac{4}{2} = -3,5.\)
Ответ: \(-3,5.\)