$\begin{align}   & y'=1-\sqrt{2}\sin x.\text{ V }\!\!\acute{\mathrm{i}}\!\!\text{ i }x\in \left[ 0;\frac{\pi }{2} \right],y'=0\Leftrightarrow \sin x=\frac{1}{\sqrt{2}} \\  & \text{C }\!\!\tilde{\mathrm{a}}\!\!\text{  }y(0)=\sqrt{2};y\left( \frac{\pi }{4} \right)=\frac{\pi }{4}+1;y\left( \frac{\pi }{2} \right)=\frac{\pi }{2}\Rightarrow M=\frac{\pi }{4}+1;m=\sqrt{2} \\ \end{align}$