Answer:
Part a)
![T_L = 155.4 N](https://img.qammunity.org/2020/formulas/physics/high-school/nbjtbl4z8azp9xtxao1pw6t8u73dvxunrz.png)
Part b)
![T_R = 379 N](https://img.qammunity.org/2020/formulas/physics/high-school/zxzpyzc2awcv5he8cku0w0pxz0awyja6tj.png)
Step-by-step explanation:
As we know that mountain climber is at rest so net force on it must be zero
So we will have force balance in X direction
![T_L cos65 = T_R cos80](https://img.qammunity.org/2020/formulas/physics/high-school/151fypvr0zl4pm50rztyib4g02efljrhc3.png)
![T_L = 0.41 T_R](https://img.qammunity.org/2020/formulas/physics/high-school/n1m14220kex84qpgtbvhkgq4789h70o6gn.png)
now we will have force balance in Y direction
![mg = T_L sin65 + T_Rsin80](https://img.qammunity.org/2020/formulas/physics/high-school/j2dk4n5xwznimbfq5tju3q2gs6vyohcpmy.png)
![514 = 0.906T_L + 0.985T_R](https://img.qammunity.org/2020/formulas/physics/high-school/37737zjllocnex7t5c1jhq3086vlfqc6nu.png)
Part a)
so from above equations we have
![514 = 0.906T_L + 0.985((T_L)/(0.41))](https://img.qammunity.org/2020/formulas/physics/high-school/jceoyhbvmzrym0knq7vazc24ok6qu0insc.png)
![514 = 3.3 T_L](https://img.qammunity.org/2020/formulas/physics/high-school/72groe3bs7o6zmq47anmrkes8zx55otrru.png)
![T_L = 155.4 N](https://img.qammunity.org/2020/formulas/physics/high-school/nbjtbl4z8azp9xtxao1pw6t8u73dvxunrz.png)
Part b)
Now for tension in right string we will have
![T_R = (T_L)/(0.41)](https://img.qammunity.org/2020/formulas/physics/high-school/d0e6h0o2r0puoa2124f70z80fb0rq5o92z.png)
![T_R = 379 N](https://img.qammunity.org/2020/formulas/physics/high-school/zxzpyzc2awcv5he8cku0w0pxz0awyja6tj.png)