a)

$\sin \alpha +\cos \alpha =\frac{1}{\sqrt{2}}\mid {\text{()}}^2$  

${\left(\sin \alpha +\cos \alpha \right)}^2=\frac{1}{2}$  

${\sin }^2\alpha +2\sin \alpha \cdot \cos \alpha +{\cos }^2\alpha =\frac{1}{2}$   

$\underset{=1}{\underbrace{{\sin }^2\alpha +{\cos }^2\alpha }}+2\sin \alpha \cdot \cos \alpha =\frac{1}{2}$  

$1+2\sin \alpha \cdot \cos \alpha =\frac{1}{2}$  

$2\sin \alpha \cdot \cos \alpha =-\frac{1}{2}\mid :2$  

$\sin \alpha \cdot \cos \alpha =-\frac{1}{4}$  

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b)

Korzystając z obliczeń z podpunktu a) mamy 

${\left(\cos \alpha -\sin \alpha \right)}^2={\cos }^2\alpha -2\cos \alpha \cdot \sin \alpha +{\sin }^2\alpha =$

$=\underset{=1}{\underbrace{{\sin }^2\alpha +{\cos }^2\alpha }}-2\cdot \cos \alpha \sin \alpha =1-2\cdot \left(-\frac{1}{4}\right)=1+\frac{1}{2}=\frac{3}{2}$  

czyli 

${\left(\sin \alpha -\cos \alpha \right)}^2=\frac{3}{2}$  

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c)

Zauważmy, że 

${\left({\sin }^2\alpha +{\cos }^2\alpha \right)}^2={\sin }^4\alpha +2\cdot {\sin }^2\alpha \cdot {\cos }^2\alpha +{\cos }^4\alpha$  

skąd dostajemy 

${\sin }^4\alpha +{\cos }^4\alpha ={\left({\sin }^2\alpha +{\cos }^2\alpha \right)}^2-2\cdot {\sin }^2\alpha \cdot {\cos }^2\alpha =$  

Korzystając z jedynki trygonometrycznej i obliczeń z podpunktu a) mamy 

$=1^2-2\cdot {\left(\sin \alpha \cdot \cos \alpha \right)}^2=1-2\cdot {\left(-\frac{1}{4}\right)}^2=1-2\cdot \frac{1}{16}=1-\frac{1}{8}=\frac{7}{8}$  

czyli 

${\sin }^4\alpha +{\cos }^4\alpha =\frac{7}{8}$