Answer
The speed of the command module is $~~4628~km/h$
Work Step by Step
We can use conservation of momentum to find the final speed of the command module. To do this we need to know the speed of the motor relative to the earth,
$$v_{me} = v_{re} + v_{mr}$$
Where $v_{re}$ is the velocity of the rocket relative to the earth ($4300 \frac{km}{h}$), $v_{mr}$ is the velocity of the motor relative to the rocket($-82 \frac{km}{h}$) because it's in the opposite direction, and $v_{me}$ is the velocity of the motor relative to earth.
$$v_{me} = 4300 \frac{km}{h} + (-82 \frac{km}{h}) = 4218 \frac{km}{h}$$
Next we can get our equations of momentum for the system before and after the departure,
$$p_i = v_{re}(4m+m) = 5mv_{re}$$
$$p_f = mv_{ce} +4mv_{me} $$
Where v_{ce} is the velocity of the command module. Substituting for v_{me},
$$p_f = mv_{ce}+4m(v_{re} + v_{mr})$$
We can than set up our conservation of momentum,
$$p_f = p_i$$
$$5mv_{re} =mv_{ce} + 4m(v_{re} + v_{mr} )$$
Substituting
$$(4m)(4218~km/h)+(m)(v_{ce}) = (5m)(4300~km/h)$$
$$(4)(4218~km/h)+v_{ce} = (5)(4300~km/h)$$
$$v_{ce} = (5)(4300~km/h)-(4)(4218~km/h)$$
$$v_{ce} = 4628~km/h$$
The speed of the command module is $~~4628~km/h$
