$\Sigma M_F = 0$

$12R_A = 4(360)$

$R_A = 120 \, \text{kN}$

$\Sigma M_C = 0$

$3F_{BD} = 4(120)$

$F_{BD} = 160 \, \text{kN}$ 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*