Since the variables are not shown I will use sin(x)=-12/13, making cos(x)=5/13 (because cos(x)=√(1-sin2(x))=√(1-(12/13)2)=-5/13, and x is in Q3), and tan(x)=12/5; and tan(y)=½, making sin(y)=-1/√5 or -√5/5, and cos(y)=-2/√5 or -2√5/5.
(a) sin(2x)=2sin(x)cos(x)=2(-12/13)(-5/13)=120/169, assuming the variable is x;
sin(2y)=2(-1/√5)(-2/√5)=⅘, assuming the variable is y. This is the more probable answer, because of the simplicity of the fraction.
(b) cos(x+y)=cos(x)cos(y)-sin(x)sin(y)=
(-5/13)(-2/√5)-(-12/13)(-√5/5)=2√5/13-12√5/65=(10√5-12√5)/65=-2√5/65.
(c) Assuming variable is x:
tan(x)=2tan(x/2)/(1-tan2(x/2)). Let t=tan(x/2), then 12/5=2t/(1-t2), 12(1-t2)=10t, 12t2+10t-12=0, 6t2+5t-6=0=(3t-2)(2t+3).
Therefore, tan(x/2)=⅔ or -3/2. Since π<x<3π/2, π/2<x/2<3π/4 (Q2), making tan(x/2)=-3/2.