The gravitational acceleration at the surface is $g = GM/R^2$, so $GM = gR^2$.

[Q1] The escape velocity $v_e$ is the speed at which the kinetic energy equals the gravitational potential energy magnitude.

$$\frac{1}{2}mv_e^2 = \frac{GMm}{R}$$ $$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} = \sqrt{2(1.0 \text{ m/s}^2)(500 \times 10^3 \text{ m})} = 1000 \text{ m/s}$$

[Q2] The initial velocity $v_0 = 1000$ m/s is equal to the calculated escape velocity. An object launched with escape velocity will theoretically travel to an infinite distance from the planet.

[Q3] Using conservation of energy for an object falling from height $h_0$. The initial position is $r_i = R+h_0$ with $v_i=0$. The final position is $r_f=R$ with speed $v_f$.

$$E_i = U_i = -\frac{GMm}{R+h_0}$$ $$E_f = K_f + U_f = \frac{1}{2}mv_f^2 - \frac{GMm}{R}$$

Setting $E_i = E_f$ and substituting $GM = gR^2$:

$$-\frac{gR^2m}{R+h_0} = \frac{1}{2}mv_f^2 - \frac{gR^2m}{R}$$ $$v_f = \sqrt{2gR^2\left(\frac{1}{R} - \frac{1}{R+h_0}\right)} = \sqrt{\frac{2gRh_0}{R+h_0}}$$

With $R = 500$ km and $h_0 = 1000$ km:

$$v_f = \sqrt{\frac{2(1.0 \text{ m/s}^2)(5 \times 10^5 \text{ m})(10 \times 10^5 \text{ m})}{5 \times 10^5 \text{ m} + 10 \times 10^5 \text{ m}}} \approx 816 \text{ m/s}$$