(2)this one is also not sufficient.i think u must know either of their speed to answer that.
i solved it mathematically.this is what i came up with.
let speed of rene = x km/min
speed of eva= y
time taken by eva=10/y
timetaken by rene=10/y+40=10/x
hence 4xy=x-y
now when rene has already reached the top,distance covered by eva=speed*time=y*10/x=10y/x
and now rene takes 10/2x time to reach at the bottom.
so eva has 10/2x to climb to top and then descent whatec\ver distance she can.
while climbing up,distace left frr eva=10-10y/x
speed=y
therefore time taken by eva to climb up = (10-10y/x)/y =(10x-10y)/xy
so now time remaining for eva while she comes down the slope= (10/2x) -(10x-10y)/xy=(15y-10x)/xy
speed =2y
hence distance travelled by eva in this much time =(15y-10x)/xy *2y=(30y-20x)/x
right now rene is at the bottom,so the distance between the two =
10-(30y-20x)/x=(30x-30y)/x=30(x-y)/x
we can substitute x-y=4xy as derived earlier on,
we get,
distance between the two=30(x-y)/x=30*4xy/x=120y
since y isstill unknown,we can not determine the distance between the two.
i hope this is not the solution or this type doesn't come in actual gmat!