There are two forces acting on the book: the force of gravity $F_g$ acting downwards and the force of the arm $F_{hb}$ acting upwards. The free body diagram is shown in the figure.
Care should be taken when determining interaction pairs. these forces although they are of the $textbf{same amount and opposite direction}$ these two forces are $textbf{not an interaction pair}$ for the reason that these two forces are the result of the action of $textbf{two different bodies}$ on the book ($F_g$ -Earth and $F_{hb}$ hand).
$textbf{The interaction pairs of forces are: $ F_ {hb}$ and $F_ {bh} $}$ the force with which the hand acts on the book and the force with which the book acts on the hand and $textbf{$ F_g$ and $F_ {bE} $}$ the force of gravity with which the Earth attracts the book and the force with which the book attracts the Earth .

Forces acting on book are force of gravity $F_g$ and force of hand $F_{hb}$. Their interaction pairs are force of book on Earth $F_{bE}$ and force of book acting on hand $F_{bh}$.

If we lower the book at an accelerating rate. The force of the hand on the book $F_ {hb}$ will decrease, so will the force of the book on the hand $F_ {bh}$, which is also logical because in order for the book to start accelerating downwards the total force needs to be looked in that direction and we can only do this by reducing the force acting upwards and according to Newton’s third law this means that its interaction pair will also be reduced.

Forces $F_ {hb}$ and $F_ {bh}$ will decrease.

In this problem we will calculate the tensions of the ropes on which the blocks are suspended.

**Known:**
$$begin{align*}
m&=5 mathrm{kg} \
g&=9.8 mathrm{m/s^2}
end{align*}$$
**Unknown:**
– The tension force in rope between ceiling and first block, $F_{t1}$
– The tension force between first and second block, $F_{t2}$

Let’s start with the tension of the rope that held the bottom block. The first thing to consider is what forces are acting on that block. The block is acted upon by the force of gravity $F_ {g2}$ and the force of tension $F_ {t2}$. Since the block is at rest, these forces are equal.

$$begin{align*}
F_g&=mcdot g \
\
F_{g2}&=mcdot g \
&=5 mathrm{kg}cdot 9.8 mathrm{m/s^2} \
&=49 mathrm{N} \
\
F_{t2}&=F_{g2} \
&=boxed{49 mathrm{N}}
end{align*}$$

We calculate the tension of the rope on which the first block hangs a little differently. Let us first consider what forces act on that block. The first block is also affected by gravity and the tension of the rope upwards, but we must not forget the tension of the rope on which the second block hangs acting on first block in downward direction. So there are three forces acting on the first block. From this we conclude that the rope tension $F_ {t1}$ is equal to the sum of the gravitational force and the rope tension $F_{t2}$.

$$begin{align*}
F_{t1}&=F_{g1}+F_{t2} \
&=mg+F_{t2} \
&=49 mathrm{N}+49 mathrm{N} \
&=boxed{98 mathrm{N}}
end{align*}$$
Note that the tension force of the rope is $F_{t1} = 2 cdot F_{t2}$.

$$begin{align*}
F_{t1}&=98 mathrm{N} \
F_{t2}&=49 mathrm{N}
end{align*}$$

Known values:

$$
begin{align*}
F_{T1}&=63 mathrm{N} \
m_2&=3 mathrm{kg} \
m_1&=?
end{align*}
$$

From the last problem we know that the tension of the rope $F_ {T2}$ on which the second block hangs is equal to its weight $F_ {g2}$, while the tension force of the rope $F_ {T1}$ on which the first block hangs is equal to the sum of its weight $F_ {g1}$ and rope tension $F_ {T2}$.

$$
begin{align*}
F_{T2}&=F_{g2} \
&=m_2cdot g \
&=3 mathrm{kg}cdot 9.8 mathrm{m/s^2} \
&=29.4 mathrm{N} \
\
F_{T1}&=F_{g1}+F_{T2} \
F_{T1}&=m_1cdot g+F_{T2} \
m_1&=dfrac{F_{T1}-F_{T2}}{g} \
&=dfrac{63 mathrm{N}-29.4 mathrm{N}}{9.8 mathrm{m/s^2}} \
&=boxed{3.429 mathrm{kg}}
end{align*}
$$

$$
m_1=3.429 mathrm{kg}
$$

In this problem we will calculate force exerted on platform by Stephanie.

Known:

$$
begin{align*}
m_b&=13 mathrm{kg} \
m_S&=61 mathrm{kg}
end{align*}
$$

Unknown:

$$
begin{align*}
g&=9.8 mathrm{m/s^2} \
F_p&=? tag{force exerted on platform}
end{align*}
$$

In order for the box that Poloma handed over to Stephanie to rest, Stephanie must act on that box with a force equal to the weight of the box. Since Stephanie acts on the box, the box acts on Stephanie with equal force. Stephanie now acts on the platform not only by her weight but by the weight of the box as well.

$$
begin{align*}
F_p&=F_{gS}+F_{gB} \
&=m_Sg+m_Bg \
&=61 mathrm{kg}cdot 9.8 mathrm{m/s^2}+13 mathrm{kg}cdot 9.8 mathrm{m/s^2} \
&=boxed{725.2 mathrm{N}}
end{align*}
$$

$$
F_p=725.2 mathrm{N}
$$

If we assume that the tree is stationary then the rope and team must be stationary. If the team acts on the rope, then the rope must act on that force of equal amount and opposite direction, and that is exactly the force of tension of the rope. So we conclude that the force of tension of the rope is equal to the force with which the team pulls the rope $F_T = F_{pull} = 500 mathrm{N}$. This problem is equivalent to the tension of the rope on which the weight is suspended from $50$ kg, because on one side of the rope is a fixed ceiling and on the other side is a weight acting on the rope with a force $F = 500 mathrm {N}$.

$$
F_T=500 mathrm{N}
$$