$$
begin{align*}
l&=3,,rm{m}\
m_1&=25,,rm{kg}\
M_2&=15,,rm{kg}
end{align*}
$$

First, lets write down torque equation around pivot point:

$$
begin{align*}
m_1gr_1&=m_2gr_2\
m_1r_1&=m_2(l-r_1)\
r_1cdot (m_1+m_2)&=m_2l\
r_1&=frac{m_2l}{m_1+m_2}\
r_1&=frac{15cdot 3}{25+15}
end{align*}
$$

$$
boxed{r_1=1.1,,rm{m}}
$$

$$
begin{align*}
F&=60,,rm{N}\
l&=1,,rm{m}\
theta &=30,,rm{^{o}}
end{align*}
$$

Torque exerted on the level can be calculated as:

$$
begin{align*}
tau &=Fcdot l cdot sin (theta)\
tau &=60cdot 1 cdot sin (30)
end{align*}
$$

$$
boxed{tau=30,,rm{N}}
$$

$$
begin{align*}
tau_{req}&=10,,rm{Nm}\
F_{max}&=50,,rm{N}
end{align*}
$$

Minimum lenght required is being used with maximum force and can be calculated as:

$$
begin{align*}
tau_{req}&=F_{max}l_{min}\
l_{min}&=frac{tau_{req}}{F_{max}}\
l_{min}&=frac{10}{50}
end{align*}
$$

$$
boxed{l_{min}=0.2,,rm{m}}
$$

$$
begin{align*}
s&=420,,rm{m}\
d&=0.42,,rm{m}
end{align*}
$$

Number of revolutions a car makes during that distance is:

$$
begin{align*}
s&=ndpi\
n&=frac{s}{dpi}\
n&=frac{420}{0.42cdot pi}
end{align*}
$$

$$
boxed{n=frac{1cdot 10^{3}}{pi},,rm{rev}}
$$

$$
begin{align*}
m&=5,,rm{kg}\
F&=25,,rm{N}\
r&=2,,rm{m}
end{align*}
$$

In order to calculate its angular acceleration, first we need to calculate its moment of inertia:

$$
begin{align*}
I&=mr^2\
I&=5cdot 2^2\
I&=20,,rm{kgm^2}
end{align*}
$$

Now we need to calculate the torque of the force:

$$
begin{align*}
tau &=Fr\
tau &=25cdot 2\
tau &=50,,rm{Nm}
end{align*}
$$

Finally we can calculate its angular acceleration:

$$
begin{align*}
alpha &=frac{tau}{I}\
alpha &=frac{50}{20}
end{align*}
$$

$$
boxed{alpha =2.5,,rm{rad/s^2}}
$$

$$
begin{align*}
d&=1.5,,rm{m}\
v&=3,,rm{m/s}
end{align*}
$$

Angular velocity can be calculated as:

$$
begin{align*}
omega &=frac{v}{0.5cdot d}\
omega &=frac{3}{0.5cdot 1.5}
end{align*}
$$

$$
boxed{omega=4,,rm{rad/s}}
$$

$$
begin{align*}
l&=0.25,,rm{m}\
F&=200,,rm{N}\
theta &=30,,rm{^{o}}
end{align*}
$$

Torque on the wrench can be calculated by multiplying:the component of the force vertical to the lever with the length of the lever:

$$
begin{align*}
tau &=lcdot Fcdot cos (theta )\
tau &=0.25cdot 200cdot cos (39)
end{align*}
$$

$$
boxed{tau =44,,rm{Nm}}
$$