Verified answer

8.

Вектора

СA ( -2,5 ; 1 )

CB ( -3,5; -1,5)

сos ( ACB ) = | CA*CB | / | CA | / | CB | = | 2,5*3,5 - 1*1,5 | / √(2,5^2+1^2) / √(3,5^2 + 1,5^2 ) = √2/2

Угол ACB = 45°

9.

Вектора

СA ( -1,5 ; -3,5 )

CB ( -3,5 ; - 1,5)

сos ( ACB ) = | CA*CB | / | CA | / | CB | = | 1,5*3,5 +  3,5*1,5 | / √(1,5^2+3,5^2) / √(3,5^2 + 1,5^2 ) = 21/29

Угол ACB = arccos ( 21/29 )  =~ 43,6°