Answer
$ \vec{F_2}=(0.59KN/m)\hat x+(1.4KN)\hat y$
$ \vec{F_1}=(-0.59KN/m)\hat x+0.33KN\hat y$
Work Step by Step
We know that
$\vec{F_2}=(\frac{m_pg}{d})x+\frac{m_bgL}{2d}$
We plug in the known values to obtain:
$\vec{F_2}=\frac{(90.0Kg)(9.8m/s^2)}{1.50m}+\frac{(85Kg)(9.81m/s^2)(5.00m)}{2(1.50m)}$
$\implies \vec{F_2}=588.6x+1389.75$
$\implies \vec{F_2}=(0.59KN/m)\hat x+(1.4KN)\hat y$
Now $\vec{F_1}=(m_b+m_p)g-F_2$
We plug in the known values to obtain:
$\vec{F_1}=(85Kg+90.0Kg)(9.81m/s^2)-(588.6x+1389.75N)$
$\implies \vec{F_1}=1716.75-588.6x-1389.75$
$\implies \vec{F_1}=(-0.59KN/m)\hat x+0.33KN\hat y$
