From FBD of beam CD
$\Sigma F_V = 0$
$R_C + R_3 = P$
$R_C + 0.5R_C = 960$
$R_C = 640 \, \text{ lb}$
$R_3 = 0.5(640) = 320 \, \text{ lb}$ answer
$\Sigma M_C = 0$
$12R_3 = 960x$
$12(320) = 960x$
$x = 4 \, \text{ ft}$
Thus, P is 8 ft to the left of D. answer
From the figure above, Rc is at the midspan of AB to produce equal reactions R1 and R2. Thus, R2 and R3 are 6 ft apart. answer
From FBD of beam AB
$R_1 = 0.5(640) = 320 \, \text{ lb}$ answer
$R_2 = 0.5(640) = 320 \, \text{ lb}$ answer