$\Sigma M_F = 0$
$12R_A = 4(360)$
$R_A = 120 \, \text{kN}$
$\Sigma M_C = 0$
$3F_{BD} = 4(120)$
$F_{BD} = 160 \, \text{kN}$ compression answer
$\Sigma F_V = 0$
$\frac{3}{5}F_{CD} = 120$
$F_{CD} = 200 \, \text{kN}$ compression answer
$\Sigma M_D = 0$
$3F_{CE} = 8(120)$
$F_{CE} = 320 \, \text{kN}$ tension answer