(N/A) Given forces acting at point $P$ are $1 \text{ N}$ at $45^{\circ}$ to the vertical,$2 \text{ N}$ at $45^{\circ}$ to the vertical,$F_2$ along the horizontal,and $F_1$ along the vertical downward.
For the system to be in equilibrium,the net force must be zero: $\sum \overrightarrow{F} = 0$.
Resolving forces along the horizontal ($x$-axis) and vertical ($y$-axis) directions:
$\sum F_x = 0 \implies F_2 + 1 \cos 45^{\circ} - 2 \cos 45^{\circ} = 0$
$F_2 + \frac{1}{\sqrt{2}} - \frac{2}{\sqrt{2}} = 0$
$F_2 - \frac{1}{\sqrt{2}} = 0 \implies F_2 = \frac{1}{\sqrt{2}} \text{ N} \approx 0.707 \text{ N}$.
$\sum F_y = 0 \implies 1 \sin 45^{\circ} + 2 \sin 45^{\circ} - F_1 = 0$
$F_1 = 3 \sin 45^{\circ} = 3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}} \text{ N} \approx 2.121 \text{ N}$.