$\cos {130}^{\circ }=\cos \left({90}^{\circ }+{40}^{\circ }\right)=-\sin {40}^{\circ }$  

 

$\sin {395}^{\circ }=\sin \left({360}^{\circ }+{35}^{\circ }\right)=\sin {35}^{\circ }$  

 

$\cos {35}^{\circ }\cdot \cos {40}^{\circ }+\left(-\sin {40}^{\circ }\right)\cdot \sin {35}^{\circ }=\cos {35}^{\circ }\cdot \cos {40}^{\circ }-\sin {40}^{\circ }\cdot \sin {35}^{\circ }$   

 

Skorzystajmy ze wzoru na cosinus sumy kątów:

$\cos \left(\alpha +\beta \right)=\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta$  

 

$\cos {35}^{\circ }\cdot \cos {40}^{\circ }-\sin {40}^{\circ }\cdot \sin {35}^{\circ }=\cos \left({35}^{\circ }+{40}^{\circ }\right)=\cos {75}^{\circ }$  

 

Skorzystajmy ponownie z tego wzoru:

$\cos {75}^{\circ }=\cos \left({30}^{\circ }+{45}^{\circ }\right)=\cos {30}^{\circ }\cdot \cos {45}^{\circ }-\sin {30}^{\circ }\cdot \sin {45}^{\circ }=\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}-\frac{1}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\underline{\frac{\sqrt{6}-\sqrt{2}}{4}}$