Calculate the electrostatic force between two spheres with charges q1 = 2 x 10⁻⁷ C and q2 = 3 x 10⁻⁷ C, separated by 30 cm.

The charges on the spheres are $q_1 = 2 \times 10^{-7} \text{ C}$ and $q_2 = 3 \times 10^{-7} \text{ C}$.
The distance between the spheres is $r = 30 \text{ cm} = 0.30 \text{ m}$.
The electrostatic constant for air is $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.

The formula for Coulomb's Law is $F = k \frac{q_1 q_2}{r^2}$.

Since both charges are positive, the force is repulsive.
The force is $\boxed{\mathbf{6 \times 10^{-3} \text{ N}}}$ and is repulsive.

The charge on the first sphere is $q_1 = 0.4 \mu\text{C} = 0.4 \times 10^{-6} \text{ C}$.
The force experienced is $F = 0.2 \text{ N}$.
The charge on the second sphere is $q_2 = 0.8 \mu\text{C} = 0.8 \times 10^{-6} \text{ C}$.
The electrostatic constant for air is $k = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.

Using Coulomb's Law, $F = k \frac{q_1 q_2}{r^2}$. We need to find $r$.
Rearranging the formula for $r^2$:

Substitute the given values:

Now, take the square root to find $r$:

Converting to centimeters:

The distance between the spheres is $\boxed{\mathbf{0.12 \text{ m} \text{ or } 12 \text{ cm}}}$.

According to Newton's third law, the force exerted by the first sphere on the second sphere is equal in magnitude and opposite in direction to the force exerted by the second sphere on the first sphere.
Therefore, if the first sphere experiences a force of $0.2 \text{ N}$ due to the second sphere, the second sphere will also experience a force of $0.2 \text{ N}$ due to the first sphere.
The force on the second sphere due to the first sphere is $\boxed{\mathbf{0.2 \text{ N}}}$.