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Make a FBD of the weight with the 2 cables cut. Sum forces in the horiz and vert directions to get the force in each cable.

http://www.lmnoeng.com/HazenWilliamsDesign.htm

You could even reduce the last problem down to a single step, your final equation F3=2.4m(170kg)(9.8)/0.8m can be gotten directly by summing moments about the top left corner of picture, where F1 and F2 intersect.

OK, I take the "contact forces" to mean you have a single horizontal reaction at the ridge and a single vertical reaction at the connection to the wall. I would not think any joints to be moment resisting in this problem. The snow should be modeled as a single concentrated load at the...

The first part is right about AE but you can't assume EF=Ex because there is also member BE framing into the joint you have to analyze.

You're doing too much work. You can solve for CD by just balancing the free body of member ABC now that you have the reactions at A. CD is a 2 force force member so it only has axial force (in its X direction), but you still have to apply the 10,000lb load to joint C in the Y direction when...

You basically have it. Your equation 1500*5-W(4-x)=0 is right and the other is just 1000*5-W*x=0 (if x is measured to the right of C) Now you need to solve the 2 equations and 2 unknowns.