How do i solve for tension on T1 and T2? ?

Resolve the forces into x and y components and balance, assuming nothing is moving

T1 cos(41°) = T2 cos(63°)

T3 = T1 sin(41°) + T2 sin(63°)

T3 = w = 200N

2 equations and 2 unknowns...

T1 = T2 cos(63°) / cos(41°)

200 = [T2 cos(63°) / cos(41°)] sin(41°) + T2 sin(63°)

200 = T2 [cos(63°) sin(41°)/ cos(41°) + sin(63°)]

T2 = 156 N ✔

T1 = 156 (cos(63°) / cos(41°))

T1 = 93.8 N ✔

Is that weight force 200 N. I cannot read it clearly. I am going to go with 200 N. Separate the three tension forces into their horizontal and vertical components.

T₁ = -|T₁|cos(41°)i + |T₁|sin(41°)j

T₂ = |T₂|cos(63°)i + |T₂|sin(63°)j

T₃ = (-200 N)j

T₁ + T₂ + T₃ = 0

When you add the component parts it comes to this system with unknowns |T₁| and |T₂|:

-|T₁|cos(41°) + |T₂|cos(63°)

|T₁|sin(41°) + |T₂|sin(63°) - 200 N = 0

Since the light isn't moving, the sum of the forces acting upward must equal the sum of the forces acting downward.

T1 * sin angle + T2 * sin angle = W

also, since the light isn't moving side to side, the sum of the horizontal forces equal to zero

T1 * cos angle = T2 * cos angle

2 equations, 2 unknowns. the math is up to you.