solve

tan2α=2tanα/( 1-tan^2α)=(-2/3)/( 1- 1/9)=-3/4

tan(2α+β)=(tan 2α+tanβ)/( 1-tan 2α* tanβ)=(-3/4- 1/7)/( 1-(-3/4)*(- 1/7))

=(-25/28)/( 1-3/28)

=- 1

Because 0

So π/2