\(\sin \alpha -\cos \alpha = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {\sin \alpha -\cos \alpha } \right)^2} = {\left( {\dfrac{1}{2}} \right)^2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha -2\sin \alpha \cos \alpha + {\cos ^2}\alpha = \dfrac{1}{4}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\)\( \Leftrightarrow \,\,\,\,\,\,\,2\sin \alpha \cos \alpha = 1-\dfrac{1}{4}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin \alpha \cos \alpha = \dfrac{3}{8}.\)
Тогда:
\(16\left( {{{\sin }^4}\alpha + {{\cos }^4}\alpha } \right) = 16\left( {{{\left( {{{\sin }^2}\alpha } \right)}^2} + 2{{\sin }^2}\alpha {{\cos }^2}\alpha + {{\left( {{{\cos }^2}\alpha } \right)}^2}-2{{\sin }^2}\alpha {{\cos }^2}\alpha } \right) = \)
\( = 16\left( {{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)}^2}-2{{\left( {\sin \alpha \cos \alpha } \right)}^2}} \right) = 16 \cdot \left( {1-2 \cdot \dfrac{9}{{64}}} \right) = 16-\dfrac{9}{2} = 16-4,5 = 11,5.\)
Ответ: 11,5.