Answer
The tension in the farther cable is $~~529.2~N$
Work Step by Step
In part (a), we found that the tension in the nearer cable is $842.8~N$
We can use the vertical forces to find the tension $T_f$ in the farther cable:
$F_{net,y} = 0$
$T_f+842.8~N - (80~kg)(9.8~m/s^2) - (60~kg)(9.8~m/s^2) = 0$
$T_f = (80~kg)(9.8~m/s^2) + (60~kg)(9.8~m/s^2) - 842.8~N$
$T_f = 529.2~N$
The tension in the farther cable is $~~529.2~N$
