Please refer to the figure showing the free body diagrams. 
We apply the law of conservation of momentum wherein the summation of momentum equals zero. Momentum is force times distance. Force equals mass times gravity, if mass is known. g=9.8 m/s2

Note: I will not indicate the units. Since all units are uniform and in SI, the answer would also be in SI units. 

1. What is the force the left support exerts on the beam?

Take the right support to be the fixed point (F3=F4=0). The equation would become

-44(9.8)(5/3 + 5/3) + F4(5/3)-117(9.8)( frac{5}{3} / 2)=0

F2 = 945.7 N

2. What is the force the right support exerts on the beam? 

Take the point where the gymnast is standing (F1=0) to be the fixed point. The equation would become

(945.7)(5/3)-(117)(9.8)(( frac{5}{3} /2)+5/3)+F4(5/3+5/3)=0

F4 = 387.1 N

3. How much extra mass could the gymnast hold before the beam begins to tip? Now the gymnast (not holding any additional mass) walks directly above the right support.

Take the the left beam to be the fixed point (F2=0). The equation would become

(-117)(9.8)( frac{5}{3} /2)-44(9.8)(5/3)+F4(5/3)=0

F4 = 1004.5 N,   this should be equal to the force exerted by the man so the beam won’t tip.

The force exerted by man is,

F1 = (44)(9.8) = 431.2 N

The excess force the gymnast could carry would be 1004.5 N – 431.2 N = 573.3 N. Converting this to mass,

Extra mass = 573.3 N ÷ 9.8
Extra mass = 58.5 kg