From the FBD of the whole system
$\Sigma M_F = 0$
$4A_H + 12A_V + 5(300 \times 10) = 6(1000)$
$4A_H + 12A_V = -9000$
$A_H + 3A_V = -2250$ ← Equation (1)
$\Sigma M_A = 0$
$12F_V = 4F_H + 6(1000) + 9(300 \times 10)$
$12F_V - 4F_H = 33\,000$
$3F_V - F_H = 8\,250$ ← Equation (2)
$\Sigma F_H = 0$
$A_H = F_H + 300(10)$
$A_H = F_H + 3000$ ← Equation (3)
From FBD of member BC
$\Sigma M_B = 0$
$10C_H = 5(300 \times 10)$
$C_H = 1500 ~ \text{lb}$
From the FBD of member CD
$\Sigma F_H = 0$
$D_H = 1500 ~ \text{lb}$
$From the FBD of member DF
$\Sigma M_E = 0$
$4F_H = 4(1500)$
$F_H = 1500 ~ \text{lb}$
Substitute FH = 1500 lb to Equation (2)
$3F_V - 1\,500 = 8\,250$
$F_V = 3250 ~ \text{lb}$
Substitute FH = 1500 lb to Equation (3)
$A_H = 1500 + 3000$
$A_H = 4500 ~ \text{lb}$
Substitute AH = 4500 lb to Equation (1)
$4500 + 3A_V = -2250$
$A_V = -2250 ~ \text{lb}$
Answer Summary
AH = 4500 lb to the left
AV = 2250 lb downward
FH = 1500 lb to the right
FV = 3250 lb upward