Answer
See the detailed answer below.
Work Step by Step
a) Since air resistance is neglected, we can use the conservation of energy principle to find the impact speed.
$$E_{i}=E_{f}$$
$$K_i+U_{gi}=K_f+U_{gf}$$
$$\frac{1}{2}mv_i^2+\dfrac{-Gm_Em}{r_i}=\frac{1}{2}mv_f^2+\dfrac{-Gm_Em}{r_f}$$
where $m_E$ is Earth's mass and $m$ is the 1-kg object mass.
We know that the ball starts from rest, so $v_i=0$.
At the ground, $r_f=R_E$ since the distance between the center of the two objects is just the radius of the Earth, where $R_E$ is Earth's radius,.
Initially, $r_i=h+R_E$ since the distance between the center of the Earth and the object is Earth's radius plus the object's height from the ground.
$$\frac{1}{2}m(0)^2-\dfrac{Gm_Em}{R_E+h}=\frac{1}{2}mv_f^2-\dfrac{Gm_Em}{R_E}$$
$$- \dfrac{Gm_E \color{red}{\bf\not}m}{R_E+h}=\frac{1}{2} \color{red}{\bf\not}mv_f^2-\dfrac{Gm_E \color{red}{\bf\not}m}{R_E}$$
$$- \dfrac{2Gm_E }{R_E+h}= v_f^2-\dfrac{2Gm_E }{R_E}$$
Hence,
$$v_f=\sqrt{ \dfrac{-2Gm_E }{R_E+h}+ \dfrac{2Gm_E }{R_E}}$$
$$v_f=\sqrt{ -2Gm_E \left[ \dfrac{1}{R_E+h}- \dfrac{1}{R_E}\right]}$$
Plugging the known;
$$v_f=\sqrt{- 2(6.67\times 10^{-11})(5.98\times 10^{24})\left[ \dfrac{1}{(6370+500)\times 10^3}- \dfrac{1}{6370\times 10^3}\right]}$$
$$v_{f1}=3019\;\rm m/s\approx\color{red}{\bf 3.02}\;km/s$$
__________________________________
b) When the Earth is flat, we can use the same conservation of energy but the gravitational potential energy is then $mgy$.
Hence,
$$K_i+U_{gi}=K_f+U_{gf}$$
$$\frac{1}{2}mv_i^2+mgy_i=\frac{1}{2}mv_f^2+mgy_f$$
where $v_i=0$, $y_f=0$; and $y_i=h$
$$0+mgy_i=\frac{1}{2}mv_f^2+0$$
So that
$$v_f=\sqrt{2mgh}=\sqrt{2(1)(9.8)(500\times 10^3)}$$
$$v_{f2}=3130.5\;\rm m/s\approx\color{red}{\bf 3.13}\;km/s$$
__________________________________
c)
The error percentage in flat earth is given by
$${\rm Error\%}=\dfrac{v_{f2}-v_{f1}}{v_{f1}}\times 100$$
$${\rm Error\%}=\dfrac{ 3130.5-3019 }{3019}\times 100=\color{red}{\bf 3.7\%}$$
