word-for-word problem:
Block B weighs 712 N. The coefficient of static friction between block B and the table is 0.25. Find the maximum weight of block A for which block B will remain at rest.

http://www.geocities.com/zdmlz/schoolrelated/

So...

block B: F_net = F_tension - f_s

...and...

block A: F_net = F_tension - mg

How do I incorporate the junction and the given angle? Is the diagonal string pulling (in the positive direction) with force of mgsin? Do I have to break-up the diagonal string into x and y components? What am I missing?

You can write down a free body diagram for:

1. Block A
2. Block B
3. The junction of the 3 strings.

Call the tension in the horizontal string T₁, the tension in the vertical string T₂, and the tension in the diagonal string T₃.

Now for applying Newton's second law.

You got block A right, except I would say T₂ instead of F_tension, simply because you don't want to use the same variable for two different tensions.

Block B is also OK, except I would say T₁ instead of F_tension, because that is not the same tension as is acting on block B.

Now for the junction. You have forces in both the x and y directions, so you have to decompose the forces by trigonometry. You already indicated that you know how to do this. Try to write down the equations, and post them here. I'll help you through the rough spots.