If a 1000N beam of 10m is supported at one end and tied to a vertical string at the other, and a person...

who weights 200N is standing 4m from the supported end, then what is the tension in the string and the reaction force at the supported end??

You have 2 different answers. So you need a tie breaker. I'll try.

I like the method in your first answer better. I can't see where the 200((6/4)+1) came from. It doesn't even look reasonable if you try some other placements of the person. So, working with the method of your first answer but explaining it my way (and it turns out having to correct a mistake) ...

The system is in equilibrium. Therefore the clockwise (cw) torques are equal and opposite the counterclockwise (ccw) torques. But we have to agree on where the support is and where the rope is so we see the same cw/ccw relationship. Let's put the support at the left so the upward force at the support tends to rotate the beam cw. That's the same way your first answer has it.:

ΣT = 0 = -Tcw + Tccw
ΣT = 0 = -(200 N)(4 m) - (1000 N)(5 m) + (T)(10 m)

T = [(200 N)(4 m) + (1000 N)(5 m)] / 10 m
T = 580 N
So far I agree with answer #1.

Now sum the forces in the vertical direction to find the reaction force R:

ΣFy = 0 = R - 1000 N - 800 N + 580 N
That's how answer #1 starts out working to find R. Ummmm.
Where did the 800 N come from? The person's weight is 200 N. The 800 N is simply an error I think. The result with that start was R = 1220 N.

Let's analyse that. The beam and the person have total weight of 1200 N. With T = 580 N and R = 1220 N, the upward forces total 1800 N versus the 1200 N downward forces. The system is supposed to be in equilibrium. Therefore the net torque must be zero -- we have that -- and the net force must be zero. Your answer #1 blew it there.

Let me restart with the analysis of the forces in the vertical direction.

ΣFy = 0 = R - 1000 N - 200 N + 580 N
R = 1000 N + 200 N -580 N = 620 N