For starters, you will need to know the angles represented on the graph. The angle the hypotenuse makes with the 3m side is equal to the arccos(3/5). The angle the hypotenuse makes with the 4m side is equal to the arccos(4/5).

Next you need to draw your free body diagram and sum your forces. Because the system is at rest, we can sum the forces in both the x and y directions, and set them equal to zero.

EFx=T4cos(arccos4/5)-T3cos(arccos3/5)=0
EFy=T4sin(arccos4/5)+T3sin(arccos3/5)-4500=0
where T4 is the tension in the 4m rope and T3 is the tension in the 3m rope. Solve the first problem for T4, then plug that into the second.

T4(4/5)=T3(3/5)
T4=(3/4)T3
Plug that into the second equation:
((3/4)T3)(3/5)+T3(4/5)=4500
Solve for T3:
(9/20)T3+(4/5)T3=4500
(5/4)T3=4500
T3=3600N