Answer:
a)    Fa = G m2 [M / r² - m / (d-r)²]  
b) r2 = 31 10⁶ m
Explanation:
The equation of the law of universal gravitation is
 
             F = G m1 m1/ r²
The value of the gravitation constant is 6.67 10-11 N m²/kg². This force is always attractive.
Let's calculate the value of that force on the spacecraft, add the strength
Earth's force to the ship
             F1 = G M m2 / r²
The moon force ship
             F2 = G m m2 / (d-r)²
Total force is
             Fa = F1 — F2
             Fa = G M M2 / r² - G m m2 / (d-r)²
             Fa = G m2 [M / r² - m / (d-r)²]
 
This is the force on the spaceship
b)  Let's look for the point where the force is zero, for this we can see that the value of the bracket must be zero
            Fa = 0 
            [M / r² - m / (d-r)²] = 0
             M / r² = m / (d-r)²
            (d-r)² = m/M   r²
            d² -2rd + r² - m/M   r² = 0
            r² [m/M - 1] + r 2d - d² = 0
This is a second degree equation for r, we solve the to find the results.
            r = {-2d ±√[4d² - 4 [m/M -1] (-d²)]} / (2 [m/M-1])
            r = {-2d ± √ [4d² (1 + (m/M-1)]} / 2(m/M-1)
            r = {-2d ± 2d √(m/M)}  / (2(m/M-1))
            r = 2d {-1 ± √(m/M)} / 2(m/M-1)
            r = d [-1 ± √(m/M)] /  (m/M-1)
To find the explicit value we substitute the values that we can find in tables
           m = 7.36 1022 kg
           M = 5.98 1024 kg
           d = 380000 km (1000m / 1 km) = 380 10⁶ m
           r = 380 10⁶ [-1 ±√(7.36 10²² / 5.98 10²⁴)] /(7.36 10²² / 5.98 / 10²⁴ -1)
           r = 380 10⁶ [-1 - √ (1.23 10²)] / (123-1)
           r = 380 106 [-1 ± 11] / 122
   
           r1 = 380 10⁶ 10/122 = 380 10⁶ 0.08197
           r1 =  31 10⁶ m
           r2 = 380 106 [-12/122] = 380 10 6 0.09836
           r2 = -37 10⁶ m
The correct distance is the positive r2 = 31 10⁶ m
c) let's use Newton's second law, to find the acceleration in the spacecraft
   
           F = m a
           a = Fa / m2 = G m2 [M / r² - m / (d-r)²] / m2
           a = G [M/r² - m/(d-r)²]
Since we have acceleration, we can use the definition of kinematics
            a = dv / dt = dv / dr dr / dt = dv / dr v
            v dv = a dr
            v dV = G [M /r² - m /(d-r)²] dr
We integrate
             ½ (V² - Vo²) = G [M (-1 /r) -m (1 / (d-r)
We evaluate between the initial point where we can assume that the initial velocity is zero for the Xo position and the final point with velocity v at the points
             V² = 2G [M (1 / Xo - 1 /X) - m (1 / (d-X) - 1 /(d-xo)]
             V² = [-M /X -m /(d-X)] 2G + constant
We now use the definition of speed
             v = dx / dt
             dx = V dt
We substitute, perform the integral and simplify, if we can make the constant zero
              dx = √([-M / X -m / (d-X)] 2G) dt
              dx / √([-M / X -m / (d-X)] 2G = dt