1) Wall reaction at D.

2) Tension in cable BC.

3) Hinged reaction at support

Applying Lami's theorem on cylinder.

$R_D/ (\sin 135) = R_P/ (\sin 90) = 1000/ \sin 135 $

Solving equations we get,

$R_D= 1000 N $ & $R_P= 1414.21 N$

Now considering beam AB

Applying conditions of equilibrium

$\sum F_x = R_P \cos 45 - T_B \sin 45 - H_A=0 -- (1) \\ \sum F_Y = - R_P \sin 45 + T_B \cos 45 - 400 + V_A = 0 -- (II) \\ \sum MFA \cong 0 [ +ve] \\ 1414.21(1.81) + 400(3 cos 45) - TB (6) = 0 \\ TBC= 568.04 N $

Putting TBC in eqn (i) & (ii)

$HA = 598.33 N, \\ VA= 998.33 N $

So, $ RA=1163.9 N $ & $\theta= 59.060^\circ$

Reaction at D is $1000N,$ Tension in BC is $568.04 N$, and Hinge reaction at A is $1163.9 N$