Answer
(a) She must fall further $3H$ distance in order to acquire a speed of $2v$
(b) The distance $(h)$ would remain same $3H$.
Work Step by Step
(a) Let down is the positive direction
$v^{2}=u^{2}+2gh$
First case:
$v^{2}=(0)^{2}+2gH$
$v^{2}=2gH$
Second case:
$(2v)^{2}=(v)^{2}+2gh$
$4v^{2}=v^{2}+2gh$
$2gh=3v^{2}$
$2gh=3\times2gH$
$h=3H$
Hence, she must fall further $3H$ distance in order to acquire a speed of $2v$.
(b) In solution (a), it is seen that distance $(h)$ in order to acquire a speed of $2v$ does not contain gravity term $(g)$. Therefore, if this event were to occur on another planet where the acceleration due to gravity had a value other than $9.80$ $m/s^{2}$, the distance $(h)$ would remain same $3H$.
