Your first method is the correct one, as long as the starting time is measured from the simultaneous moment at $t=0$ in the Earth's frame. As you probably know, it's not very natural to talk about simultaneous moments for spatially separated events, and this is where part of the subtlety in this question lies.
By the way, your first method can be found more simply by just dividing the Earth frame time $6000s$ by the time dilation factor gamma. $$\frac{3}{5}6000s=3600s\qquad \frac{4}{5}6000s=4800s$$
Now let's go to one of the ships' frames, say the one moving at .8c. The time $t'=0$ is when that ship starts its 90 minute timer, but the other ship started its timer at some earlier $t'$!
The event at which the other ship starts its clock is at $$t'=\gamma(t-vx)=\frac{5}{3}(0-.8\times 8400 s)=-11200 s$$ $$x'=\frac{5}{3}(8400s-.8\times 0)=14000s$$ So at $t'=0$ the other ship is at $14000s-.946\times11200s\approx3405s$, which does not agree with your Lorentz contracted value. 562
Continuing on from this distance, you will find it is $3600s$ until collision again, as in your first method.
Why length contraction failed
In order to use the length contraction idea, you need to be transforming a 'proper length.' What that means is that the endpoints of the length are both at rest in the original frame. In your case the endpoints of the distance were the two ships, which were not only not at rest, they were at different velocities.
But just to show where length contraction would come in, let's consider the proper length with endpoints given by the worldlines $x=0, \,x=8400s$. We found the transformed coordinates of the point $x=8400s$ at $t=0$ above. Now in the transformed frame the $x=0$ worldline is moving with velocity $-.8c$, so at $t'=-11200s$ it is at $x'=8960s$, and $14000s-8960s=5040s$ which was the contracted length you found in the second method. You can see it has nothing to do with the distance between the two ships. 586