Static Equilibrium Problem with 3 Tension Forces 
I had some trouble solving these.
For the first one I assumed there wasn't a tension force on the 2nd mass, which gave me:
$$FT_1.\cos A = FT_2.\cos D$$ and $$FT_1\sin A + FT_2\sin D = F.g$$
Not sure how to approach 2nd problem.
 A: It sounds like you are supposed to assume the masses are all stationary, and you are to find the ratios between the masses that will do that.   You know the tension in every rope, each one is mg where m is the mass the string is attached to.  That tension is preserved all along the string, so bring it back to the point where all the strings attach.  Then say the forces all add up to zero at that central point, and that will give you the constraint on the mass ratios.  For the first one, it's obvious that you need m_1 = m_3 from symmetry, but then m_2 depends on the angle at the center.
A: These questions are to do with the equilibrium of a point mass (knot) subjected to three forces provided by three masses and transmitted via pulleys and strings to the knot.
There are many methods of solving such problems but drawing a force diagram, which will be a triangle in this case, might help?
Because you have a static equilibrium situation the sum of the forces acting on the knot is zero so the vector addition of the three forces produces a triangle.

Solution can be obtained by resolving forces or by use of the sine rule and the cosine rule for the triangle of forces. 
For your second problem you need to redraw the forces diagram which now will include the angle $f$.
