Answer
$T_1=100\;\rm N$
$T_2=T_3=T_5=F=500\;\rm N$
$T_4=150\;\rm N$
Work Step by Step
This is the easiest problem we will ever see, but we just need to draw the force diagram of the objects inside the system.
Since the system is in equilibrium, the net force exerted on any object is zero.
$\Rightarrow$ According to the first figure below,
$$\sum F_{block}=T_1-mg=ma_y=m(0)$$
Thus;
$$T_1=mg=10.2\times 9.8=\color{red}{\bf 100}\;\rm N$$
$\Rightarrow$ According to the second figure below,
$$\sum F_{pulley_1}=T_2+T_3-T_1=ma_y=(0)(0)$$
Recalling that the pulley is massless.
The tension $T_2$ and $T_3$ must be equal otherwise the pulley will rotate and we know that is not the case since the whole system is stationary and at rest.
Thus,
$$T_2+T_3=\frac{1}{2}T_1=T_1=100\;\rm N$$
So that
$$T_2=T_3=\dfrac{100}{2}=\color{red}{\bf 50}\;\rm N$$
$\Rightarrow$ According to the third figure below,
$$\sum F_{pulley_2}=T_4-T_2-T_3-T_5=ma_y=(0)(0)$$
Recalling that the pulley is massless.
The tension $T_3$ and $T_5$ must be equal otherwise the pulley will rotate and we know that is not the case since the whole system is stationary and at rest.
And since, $T_2=T_3$, so that $T_2=T_3=T_5=\frac{1}{2}T_1$
Therefore;
$$T_5=\color{red}{\bf 100}\;\rm N$$
And hence,
$$ T_4= T_2+T_3+T_5=50+50+50=\color{red}{\bf 150}\;\rm N$$
$\Rightarrow$ According to the last figure below,
$$F=T_5=\color{red}{\bf 50}\;\rm N$$
