Воспользуемся формулой косинуса двойного угла:
\(\cos 2\alpha = 1-2{\sin ^2}\alpha .\)
\(\cos \alpha = \dfrac{1}{8}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,1-2{\sin ^2}\dfrac{\alpha }{2} = \dfrac{1}{8}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\dfrac{\alpha }{2} = \dfrac{7}{{16}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \dfrac{\alpha }{2} = \dfrac{{\sqrt 7 }}{4},\,\,\,}\\{\sin d\dfrac{\alpha }{2} = -\dfrac{{\sqrt 7 }}{4}.}\end{array}} \right.\)
Так как \(\alpha \, \in \,\left( {0;\dfrac{\pi }{2}} \right)\), то \(\dfrac{\alpha }{2}\, \in \,\left( {0;\dfrac{\pi }{4}} \right)\) (I четверть), значит \(\sin \dfrac{\alpha }{2} > 0\), то есть \(\sin \dfrac{\alpha }{2} = \dfrac{{\sqrt 7 }}{4}.\)
Тогда: \(\sqrt 7 \sin \dfrac{\alpha }{2} = \sqrt 7 \cdot \dfrac{{\sqrt 7 }}{4} = \dfrac{7}{4} = 1,75.\)
Ответ: 1,75.