The weight of a body at height \( h \) is given by:

\[
W_h = W_0 \left( \frac{R}{R + h} \right)^2.
\]

To find \( h \) when \( W_h = \frac{1}{16} W_0 \):

\[
\frac{1}{16} = \left( \frac{R}{R + h} \right)^2.
\]

Taking the square root:

\[
\frac{1}{4} = \frac{R}{R + h}.
\]

Cross-multiplying gives:

\[
R + h = 4R ; h = 3R.
\]

Thus, the height is \( 3R \).

The acceleration due to gravity \( g' \) on the new planet can be calculated using the formula:

\[
g' = \frac{GM}{R^2}.
\]

For the new planet:

- Its radius \( R' = 3R \) (3 times the radius of Earth).
- Its mass \( M' \) is given by \( M' = \text{density} \times \text{volume} = \rho \times \frac{4}{3} \pi (R')^3 = \rho \times \frac{4}{3} \pi (3R)^3 = 27 \times \rho \times \frac{4}{3} \pi R^3 = 27M \) (mass is 27 times that of Earth).

Substituting into the formula:

\[
g' = \frac{G(27M)}{(3R)^2} = \frac{27GM}{9R^2} = 3g.
\]

Thus, \( g' = 3g \).

The height \( h \) at which gravity decreases by 36% (to 64% of its surface value) is found using:

\[
g_h = \frac{gR^2}{(R + h)^2}.
\]

Setting \( g_h = 0.64g \):

\[
0.64 = \frac{R^2}{(R + h)^2}.
\]

Taking the square root:

\[
0.8 = \frac{R}{R + h}.
\]

Cross-multiplying gives:

\[
0.8(R + h) = R \implies 0.8h = 0.2R \implies h = 0.25R=R/4
\]

Thus, the height is R/4.

The point where the gravitational force on the projectile is zero occurs when the gravitational forces from both planets are equal.

\[
\frac{G \cdot 4M \cdot m}{r^2} = \frac{G \cdot 9M \cdot m}{(6R - r)^2}
\]

Simplifying,

\[
\frac{4}{r^2} = \frac{9}{(6R - r)^2}
\]

Taking the square root:

\[
\frac{2}{r} = \frac{3}{6R - r}
\]

Cross-multiplying:

\[
2(6R - r) = 3r
\]

Solving:

\[
12R - 2r = 3r
\]

\[
5r = 12R
\]

\[
r = \frac{12R}{5}
\]

So, the distance from the lighter planet is \( \frac{12R}{5} \).

Gravitational potential \( V \) at the midpoint is the sum of the potentials due to both masses:

\[
V = -\frac{G \cdot 10^2}{0.5} - \frac{G \cdot 10^3}{0.5}
\]

Simplifying:

\[
V = -2G(10^2 + 10^3) = -2G \cdot 1100
\]

So, the gravitational potential at the midpoint is:

\[
V = -2200G
\]

The gravitational potential inside a spherical shell (including at the center) due to the shell is constant. The potential at a distance \( a/2 \) from the center is the sum of the potentials due to the mass at the center and the shell.

\[
V = -\frac{GM}{a/2} - \frac{GM}{a}
\]

Simplifying:

\[
V = -\frac{2GM}{a} - \frac{GM}{a} = -\frac{3GM}{a}
\]

So, the potential at \( a/2 \) is:

\[
V = -\frac{3GM}{a}
\]

The gravitational potential \( V \) at height \( h \) from the Earth's surface is given by:

\[
V = -\frac{GM}{R + h}
\]

The acceleration due to gravity \( g \) at height \( h \) is:

\[
g = \frac{GM}{(R + h)^2}
\]

Given:
- \( V = -5.4 \times 10^7 \, \text{J/kg} \)
- \( g = 6.0 \, \text{m/s}^2 \)
- \( R = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \)

From the first equation:

\[
-5.4 \times 10^7 = -\frac{GM}{R + h}
\]

From the second equation:

\[
6.0 = \frac{GM}{(R + h)^2}
\]

Dividing the two equations to eliminate \( GM \):

\[
\frac{V}{g} = \frac{-(R + h)}{(R + h)^2}
\]

Simplifying:

\[
\frac{-5.4 \times 10^7}{6.0} = -(R + h)
\]

Solving for \( h \):

\[
R + h = 9 \times 10^6 \, \text{m}
\]

\[
h = 9 \times 10^6 - 6.4 \times 10^6 = 2.6 \times 10^6 \, \text{m} = 2600 \, \text{km}
\]

Thus, the height is \( 2600 \, \text{km} \).

The velocity required to reach a height equal to the Earth's radius \( R \) is the escape velocity for a total distance of \( 2R \) from the Earth's center.

Escape velocity formula:

\[
v = \sqrt{\frac{GM}{R}}
\]

At height \( h = R \), the velocity needed is:

\[
v = \sqrt{\frac{GM}{2R}} = \frac{V_e}{\sqrt{2}}
\]

Thus, the velocity is:

\[
v = \frac{V_e}{\sqrt{2}} = \frac{\sqrt{2GM/R}}{\sqrt{2}} = \sqrt{\frac{GM}{R}} = V_e/\sqrt{2}= \sqrt{\frac{GM}{R}}
\]

The point where the gravitational field is zero lies closer to the smaller mass \( m \). Let the distance of this point from \( m \) be \( x \), and from \( 4m \) be \( r - x \).

At this point:

\[
\frac{Gm}{x^2} = \frac{G \cdot 4m}{(r - x)^2}
\]

Taking the square root:

\[
\frac{1}{x} = \frac{2}{r - x}
\]

Solving:

\[
r - x = 2x \quad \Rightarrow \quad x = \frac{r}{3}
\]

The gravitational potential \( V \) at this point is the sum of the potentials due to both masses:

\[
V = -\frac{Gm}{\frac{r}{3}} - \frac{G \cdot 4m}{\frac{2r}{3}} = -\frac{3Gm}{r} - \frac{6Gm}{r} = -\frac{9Gm}{r}
\]

Thus, the potential at the point is:

\[
V = -\frac{9Gm}{r}
\]

The gravitational potential at the center due to a semicircular rod is given by:

\[
V = - \frac{GM}{R}
\]

Where \( R \) is the radius of the semicircle, which is related to the length of the rod:

\[
L = \pi R \quad \Rightarrow \quad R = \frac{L}{\pi}
\]

Substituting \( R \) into the potential formula:

\[
V = - \frac{GM}{\frac{L}{\pi}} = - \frac{\pi GM}{L}
\]

Thus, the gravitational potential at the center is:

\[
V = - \frac{\pi GM}{L}
\]