Answer
a) $T=2260N$
b) $A_x=1452.7N$ and $A_y=1449N$
Work Step by Step
The forces are shown in the free-body diagram below.
Tension force $T$ has 2 components, $T_x=T\cos50$, pointing to the left, and $T_y=T\sin50$, pointing upward.
The force the wall exerts on the beam $A$ has 2 components, $A_x=A\cos30$, pointing to the right, and $A_y=A\sin30$, pointing upward.
The weight of the crate $W_c$ and of the beam $W_b$ both point downward.
Because the system is in equilibrium, the net force is zero: $$\sum F_x=A_x-T_x=0 (1)$$ $$\sum F_y=A_y+T_y-W_c-W_b=0$$ $$A_y+T_y=1960N+1220N=3180N (2)$$
Now we take the axis of rotation to be the beam's left end, which means $A$ produces no torque. Taking the beam's length to be $l$, we have $$\sum\tau=-\tau_{W_b}-\tau_{W_c}+\tau_{T}=0$$ $$-W_b(0.5l\cos30)-W_c(l\cos30)+T(l\sin80)=0$$ $$-W_b(0.5\cos30)-W_c(\cos30)+T(\sin80)=0$$ $$T=\frac{W_c\cos30+0.5W_b\cos30}{\sin80}$$
We have $W_c=1960N$ and $W_b=1220N$, so $T=2260N$
Therefore, $T_x=1452.7N$ and $T_y=1731N$
(1): $A_x=1452.7N$
(2): $A_y=3180N-1731N=1449N$
