\(\sin \alpha -\cos \alpha = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {\sin \alpha -\cos \alpha } \right)^2} = {\left( {\dfrac{1}{2}} \right)^2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha -2\sin \alpha \cos \alpha + {\cos ^2}\alpha = \dfrac{1}{4}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\)
\( \Leftrightarrow \,\,\,\,\,\,\,2\sin \alpha \cos \alpha = 1-\dfrac{1}{4}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin \alpha \cos \alpha = \dfrac{3}{8}.\)
Тогда:
\({\sin ^4}\alpha + {\cos ^4}\alpha = {\left( {{{\sin }^2}\alpha } \right)^2} + 2{\sin ^2}\alpha {\cos ^2}\alpha + {\left( {{{\cos }^2}\alpha } \right)^2}-2{\sin ^2}\alpha {\cos ^2}\alpha = \)
\( = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^2}-2{\left( {\sin \alpha \cos \alpha } \right)^2} = 1-2 \cdot \dfrac{9}{{64}} = 1-\dfrac{9}{{32}} = \dfrac{{23}}{{32}}.\)
Ответ: \(\dfrac{{23}}{{32}}.\)