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

Assuming you mean simply supported (no moment at the supported end)...

First sum the moments about the supported end to find the tension T. We'll call a counterclockwise moment positive:

ΣM = 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

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

ΣFy = 0 = R - 1000 N - 800 N + 580 N
R = 1220 N

EDIT: Whoops, made a typo, as sojsail noticed. Here's that last bit corrected:

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

Without the person standing on the beam, the string and the support are both subjected to 500N.
(10m - 4m) = 6m.
200/ ((6/4)+1) = 133.33N. force added to the support, and (200 - 133.33) = 66.67N. added to string.
Tension in string = (500 + 66.67) = 566.67N.
Reaction on supported end = (500 + 133.33) = 633.33N.

Lets call supported end T1 & tied end T2

T1:

(1000 x (10/2)) + (200 x 6) = 6200

6200 = 10T1

6200/10 = T1 = 620 N



T2:

(1000 x (10/2)) + (200 x 4) = 5800

5800 = 10T2

5800/10 = T2 = 580 N


The sum of the forces must equal:

620 + 580 = 1200 N

1000 + 200 = 1200 N

Sojsail is therefore correct.