Start off the 1 kg mass right on the end of the string. It's moving around in a circle at 5 revolutions a second -> w = 10*pi rad/s. You're given that the string rotates in the horizontal plane so the radius is r = 1m. The net force on the outer 1 kg mass (aka the centripetal force) is given by:
m*(w^2)*r = 100*pi^2 N ~ 987 N inwards.
The fact that you're given the entire string rotates in the horizontal plane indicates the absence of gravity (otherwise there would be a vertical component of force on the object, so it would be like the conical pendulum instead), so the only force acting on the string is tension (someone's probably rotating the string in outer space). So the tension force in the outer part of the string (connected to the 1 kg mass) is equal to the centripetal force, which is ~ 987 N inwards.
Now for the 2kg mass, the net (centripetal) force is m*(w^2)*r = 2*(100*pi^2)*0.5 ~ 987 N inwards. There are two tension forces acting on the inner 2 kg mass, one pulling it inwards towards the centre, and one pulling it outwards away from the centre. The vector sum of these two forces is the centripetal force of ~ 987 N inwards we just found. We know that the tension force pulling it outwards is simply equal in magnitude to the tension force pulling the 1 kg mass inwards, but in the opposite direction (action-reaction pair). So we have one force pulling the 2 kg mass outwards by 987 N. We also know that the net force on the 2 kg mass is 987 N inwards. Therefore the tension in the inner part of the string is 100*pi^2 + 100*pi^2 N ~ 1974 N inwards
Tension in outer string = 987 N inwards
Tension in inner string = 1974 N inwards
