Sections through the 2nd and 3rd panels

From the Free Body Diagram of Section Through M-M

Without the diagonals:

$\Sigma F_v = 190 - 80 = 110 ~ \text{kN upward}$
This 110 kN force must be countered by downward components of the diagonals. Member CD will be in compression. Since diagonals can support only tension, the force in member CD must be zero. The acting diagonal therefore is member BE.

$\Sigma F_v = 0$
$\frac{3}{5}F_{BE} + 80 = 190$

$F_{BE} = 183.33 ~ \text{kN}$ *answer*

From the Free Body Diagram of Section Through N-N

Without the diagonals:

$\Sigma F_v = 190 - 80 - 200 = -90 ~ \text{kN} = 90 ~ \text{kN downward}$
Since ΣF_{v} is downward, the counter diagonals must act upward. DG is in compression and therefore redundant, the acting member is the diagonal EF.

$\Sigma F_v = 0$
$\frac{3}{5}F_{EF} + 190 = 80 + 200$

$F_{EF} = 150 ~ \text{kN}$ *answer*