(1) Find the value of B between 0 degree and 360 degree which satisfy the equation SinB+Sin²B=0

(2) Find the value of A between 0 degree and 360 degree which satisfy the equation
(2CosA+1)(1-CosA)=0
(3) 2Sin²A-SinA-1=0
(4) 4Cos²A+2CosA=1
(5) 1+CosB/2-CosB=1

(1) SinB+Sin²B

sinB(1+sinB) = 0
so, sinB = 0 or -1
which angles fit that?

(2) (2CosA+1)(1-CosA)=0
cosA = 1 or -1/2

(3) 2Sin²A-SinA-1=0
(2sinA+1)(sinA-1) 0
see #2

(4) 4Cos²A+2CosA=1
(2cosA+1)^2 = 0
cosA = -1/2
...
(5) 1+CosB/2-CosB=1
1+cosB = 2-cosB
2cosB - 1 = 0
cosB = 1/2

Nothing hard about these. Just disguised exercises from Algebra I.