Воспользуемся основным тригонометрическим тождеством:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {-\dfrac{2}{{\sqrt 7 }}} \right)^2} + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\cos ^2}\alpha = \dfrac{3}{7}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\cos \alpha = \dfrac{{\sqrt 3 }}{{\sqrt 7 }},\,\,\,}\\{\cos \alpha = -\dfrac{{\sqrt 3 }}{{\sqrt 7 }}.}\end{array}} \right.\)
Так как \(\dfrac{{3\pi }}{2} < \alpha < 2\pi \) (IV четверть), то \(\cos \alpha > 0\), то есть \(\cos \alpha = \dfrac{{\sqrt 3 }}{{\sqrt 7 }}.\)
\(\sqrt {28} \cos \left( {\alpha + \dfrac{\pi }{6}} \right) = 2\sqrt 7 \left( {\cos \alpha \cos \dfrac{\pi }{6}-\sin \alpha \sin \dfrac{\pi }{6}} \right) = 2\sqrt 7 \left( {\dfrac{{\sqrt 3 }}{{\sqrt 7 }} \cdot \dfrac{{\sqrt 3 }}{2} + \dfrac{2}{{\sqrt 7 }} \cdot \dfrac{1}{2}} \right) = 3 + 2 = 5.\)
Ответ: 5.