(A) Given: $r_p = 2R$ and $r_a = 6R$.
Using the properties of an ellipse: $r_a = a(1+e) = 6R$ and $r_p = a(1-e) = 2R$.
Dividing the two equations: $\frac{1+e}{1-e} = \frac{6R}{2R} = 3 \implies 1+e = 3 - 3e \implies 4e = 2 \implies e = 0.5$.
Using conservation of angular momentum: $m v_p r_p = m v_a r_a \implies v_a = v_p \frac{r_p}{r_a} = v_p \frac{2R}{6R} = \frac{v_p}{3}$.
Using conservation of energy: $\frac{1}{2} m v_p^2 - \frac{GMm}{r_p} = \frac{1}{2} m v_a^2 - \frac{GMm}{r_a}$.
Substituting $v_a = v_p/3$: $\frac{1}{2} v_p^2 (1 - 1/9) = GM (\frac{1}{2R} - \frac{1}{6R}) = GM (\frac{2}{6R}) = \frac{GM}{3R}$.
$\frac{4}{9} v_p^2 = \frac{GM}{3R} \implies v_p^2 = \frac{3GM}{4R} \implies v_p = \sqrt{\frac{3 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{4 \times 6400 \times 10^3}} \approx 5.59 \, km/s$.
Then $v_a = v_p/3 \approx 1.86 \, km/s$.
To transfer to a circular orbit of radius $6R$,the satellite must be at apogee $(r=6R)$ and its velocity must be increased to the orbital velocity $v_c = \sqrt{\frac{GM}{6R}} \approx 2.64 \, km/s$.