solution:

This problem is not as difficult as it may seem. There will be a lot of work involved, but if you break the problem down into its parts, it is really fairly simple.
We need to find the angles phi and theta of the normal forces that assures the system equilibrium.
We start by breaking down the problem into three seperate parts, one for the block, and one for each pulley.
We start with the block.

Since the system is not moving, the sum of the forces acting on block one must add up to zero.

Now we know that the tension in the massless cable is equal to the weight of the block.
We move on to the larger pulley on the left and make a diagram with all the forces involved

With an expression for the Tension in terms m, we move on to the equilibrium states in the x and y directions.

Now we move on to the small pulley on the right.

We follow the same procedure as with the large pulley using the equilibrium condition in the x-direction of the small pulley.

Now for the equilibrium condition in the y-direction

Now we have four equations with the normal forces on pulleys one and two and the angles theta and phi.
We start with the equations for the larger pulley:

Taking the second equation we solve for , the value of which we plug into the first equation.

Now we take the two equations for the smaller pulley, this time solving for phi in a different way (you get the same answer either way, it's just a different method for solving). We divide the two equations.