$\Sigma M_A = 0$
$12R_H = 3(2.7) + 6(4.5) + 9(1.8)$
$R_H = 4.275 ~ \text{kN}$
From the section to the right of M-N
$\Sigma M_E = 0$
$6(F_{DF} \sin 30^\circ) + 3(1.8) = 6(4.275)$
$F_{DF} = 6.75 ~ \text{kN compression}$ answer
$\Sigma M_H = 0$
$6(F_{EF} \sin 30^\circ) = 3(1.8)$
$F_{EF} = 1.8 ~ \text{kN compression}$ answer
$y = 3 \tan 30^\circ = \sqrt{3}$
$\Sigma M_F = 0$
$y F_{EG} = 3(4.275)$
$\sqrt{3} F_{EG} = 3(4.275)$
$F_{EG} = 7.404 ~ \text{kN tension}$ answer