Question 202965
Problem 1. )d = distance, r = rate(speed), and t = time of travel)
For the first cyclist:
{{{d[1] = r[1]*t[1]}}}
For the second cyclist:
{{{d[2] = r[2]*t[2]}}}
The question is at what time, t, will {{{d[1] = d[2]}}}?
The first cyclist travels 3 hours longer than the second cyclist, so...
{{{t[1] = t[2]+3}}}, so rewriting the first two equations, we get:
{{{d[1] = 6(t[2]+3)}}} and
{{{d[2]  10*t[2]}}} but, when the cyclists meet, {{{d[1] = d[2]}}}, so...
{{{6(t[2]+3) = 10*t[2]}}} Solve for {{{t[2]}}}.
{{{6*t[2]+18 = 10*t[2]}}} Subtract {{{6*t[2]}}} from both sides.
{{{18 = 4*t[2]}}} Divide both sides by 4.
{{{t[2] = 4.5}}}hours.
The second cyclist will catch up with first cyclist after 4.5 hours.
The first cyclist however will have traveled for (4.5+3) = 7.5 hours. 
So if you start counting when the first cyclist begins, then the answer is 7.5 hours.
If you start counting when the second cyclist begins, then the answer is 4.5 hours.
The problem is not clear as to when you should start the clock!

Problem 2.
It's best to work out the rate (per minute) at which each of the three individuals work.
If Jim can fill the pool in 30 minutes, then he can fill 1/30 of the pool in 1 minute.
If Sue can fill the pool in 45 minutes, then she can fill 1/45 of the pool in 1 minute.
If Tony can fill the pool in 1 1/2 hours (that's 90 minutes), then he can fill 1/90 of the pool in 1 minute.
Working together, the three of them can fill (1/30 + 1/45 + 1/90 = 1/15) of the pool in 1 minute.
This means that they can fill the entire pool in 15 minutes working together.