1. If you travel 7000 m at a rate of 25 m/s and, how long does it take to complete the trip? 2. How long does it take to get back?

1.
Time taken for the trip
= Distance / Rate
= 7000/25 s
= 280 s
= 4 min 40 s

2.
If the rate is unchanged, time taken to get back
= Distance / Rate
= 7000/25 s
= 280 s
= 4 min 40 s

3.
Total time for the trip there and back
= 280 × 2 s
= 560 s
= 9 min 20 s

7000m is your 'given', start there:
7000m (1sec/25m) = 280 s
There and back just doubles the time: 560 s
((note: in KMS system, time is reported in seconds))

time for getting there = tf = 7000 m /  25 m/sec = 280 sec 
time for getting back = tr = 7000 m / 25 m/sec = 280 sec 
total trip time t = tf+tr = 280+280 = 560 sec (a bit less than 10 min) 

I must make a comment on the answer given.  How are you going to go from +25 m/s to -25 m/s without taking some time to do so?  There will be a discrepancy between what is possible and what the question asks.

7000/25=280s.
If the speed is the same 280s.
total time 560s= 9.33min.=9m 19.8s.

Do you know, I was not good at maths at school in the 1960s, but I could do that sum in my head, and write down my workings just to make sure.