In order to compute the ratio of the gravitational field of the Sun and the Earth, we have to compute the gravitational field individually and then divide them. The gravitational field in any point is given by:

$$
begin{align*}
g&=Gcdotfrac{m}{r^2}
end{align*}
$$

where the $G$ stands for the universal gravitational constant, $m$ is the mass of the planet, and the $r$ is the distance between the center of the planet and the point in which we compute the gravitational field.

The magnitude of gravitation field of the Sun in the center of the Moon:

$$
begin{align*}
g_s&=Gcdotfrac{m_s}{r^2_{sm}}
end{align*}
$$

The magnitude of gravitation field of the Earth in the center of the Moon:

$$
begin{align*}
g_e&=Gcdotfrac{m_e}{r^2_{em}}
end{align*}
$$

So the ratio will be:

$$
begin{align*}
frac{g_s}{g_e}&=frac{Gcdotfrac{m_s}{r^2_{sm}}}{Gcdotfrac{m_e}{r^2_{em}}}\
frac{g_s}{g_e}&=frac{frac{m_s}{r^2_{sm}}}{frac{m_e}{r^2_{em}}}\
frac{g_s}{g_e}&=frac{{m_s}cdot{r^2_{em}}}{{m_e}cdot{r^2_{sm}}}
end{align*}
$$

Let’s substitute:

$$
begin{align*}
frac{g_s}{g_e}&=frac{{2cdot{10^{30}}text{ kg}}cdot{(3.9cdot{10^8}text{ m})^2}}{{6cdot{10^{24}}text{ kg}}cdot{(1.5cdot{10^{11}}text{ m})^2}}
end{align*}
$$

$$
boxed{frac{g_s}{g_e}=2.253}
$$

To compute the net gravitational field magnitude, we have to compute the magnitude of the gravitational field of the Earth and Sun and add them as a vector.
The magnitude of gravitation field of the Sun in the center of the Moon:

$$
begin{align*}
g_s&=Gcdotfrac{m_s}{r^2_{sm}}\
g_s&=6.67cdot{10^{-11}} frac{text{ N}cdottext{ m}^2}{text{kg}^2}cdotfrac{2cdot{10^{30}}text{ kg}}{(1.5cdot{10^{11}}text{ m})^2}\
g_s&=5.93cdot{10^{-3}} frac{text{ N}}{text{ kg}}
end{align*}
$$

The magnitude of gravitation field of the Earth in the center of the Moon:

$$
begin{align*}
g_e&=Gcdotfrac{m_e}{r^2_{em}}\
g_e&=6.67cdot{10^{-11}} frac{text{ N}cdottext{ m}^2}{text{kg}^2}cdotfrac{6cdot{10^{24}}text{ kg}}{(3.9cdot{10^8}text{ m})^2}\
g_e&=2.63cdot{10^{-3}} frac{text{ N}}{text{ kg}}
end{align*}
$$

The net of gravitational field will be:

$$
begin{align*}
overrightarrow{g}&=overrightarrow{g_e}+overrightarrow{g_s}
end{align*}
$$

As the angle between two vectors is $90text{textdegree}$ the net magnitude of gravitational field will be:

$$
begin{align*}
g&=sqrt{g^2_s+g^2_e}\
g&=sqrt{left(5.93cdot{10^{-3}} frac{text{ N}}{text{ kg}}right)^2+left(2.63cdot{10^{-3}} frac{text{ N}}{text{ kg}}right)^2}
end{align*}
$$

$$
boxed{g=6.49cdot{10^{-3}} frac{text{ N}}{text{ kg}}}
$$

b) $g=6.49cdot{10^{-3}} frac{text{ N}}{text{ kg}}$

$$
begin{align*}
g&=Gcdotfrac{m}{r^2}
end{align*}
$$

where the $G$ stands for the universal gravitational constant, $m$ is the mass of the planet, and the $r$ is the distance between the center of the planet and the point in which we compute the gravitational field.

The magnitude of gravitation field on the moon surface will be:

$$
begin{align*}
g_m&=Gcdotfrac{m_m}{r^2_{m}}\
g_m&=6.67cdot{10^{-11}} frac{text{ N}cdottext{ m}^2}{text{kg}^2}cdotfrac{7.3cdot{10^{22}}text{ kg}}{(1.785cdot{10^6}text{ m})^2}
end{align*}
$$

$$
boxed{g_m=1.53 frac{text{ N}}{text{ kg}}}
$$

The first one is the velocity of the satellite relative to the Earth center, described by the equation:

$$
begin{align*}
v&=sqrtfrac{Gcdot{m_e}}{r}
end{align*}
$$

where the $G$ stands for the universal gravitational constant, $m_e$ is the mass of the planet that attracts the satellite (in our case the Earth), and $r$ is the distance between the radius of the orbit.
The second one is the period of satellite rotation, that is given by:

$$
begin{align*}
T&=2cdotpicdotsqrtfrac{r^3}{Gcdot{m_e}}
end{align*}
$$

In both equations there is no mass of the satellite included, so the scientists were not able to compute the mass of the satellite.

The orbital period of satellite rotation is given by:

$$
begin{align*}
T&=2cdotpicdotsqrtfrac{r^3}{Gcdot{m_e}}
end{align*}
$$

where $r$ stands for the radius of the satellite orbit.

As the radius of orbit is direct proportional to a orbital period,

$$
boxed{text{The satellite that is higher has the larger orbital period.}}
$$

$bold{b)}$

The Velocity of the satellite relative to the Earth is described by the equation:

$$
begin{align*}
v&=sqrtfrac{Gcdot{m_e}}{r}
end{align*}
$$

As the radius of the orbit is irreversible proportional to a satellite velocity,

$$
boxed{text{The satellite that is higher has a lower velocity.}}
$$

$$
begin{align*}
overrightarrow{F}=-mcdotoverrightarrow{a}
end{align*}
$$

where the $m$ stands for the mass of the chair.
We know that the chair in a spacecraft is weightless, but the reason for that is not because the chair doesn’t have the mass. The mass of the object is constant. The reason for masslessness in space is a deficiency of gravitational field.

$$
boxed{text{You will feel the pain if you kick the chair in space.}}
$$