While I would normally be the first to argue that people use too many significant figures, and leave it at that - I think there are two important points worth noting here:
First - the times are given with a precision of 1/100th of a second over almost 150 seconds - we should take that into account when formulating our answer
Second - when you say "how big can something be", it is worth considering whether rounding up is appropriate - and what the word "can" really means.
Now for the math:
If we assume that the nominal track length is $D$, nominal time $t$, nominal speed $v$, then the additional distance covered in 0.20 seconds is
$$\Delta x = v \Delta t = \frac{D}{t}\Delta t$$
The error in that measurement is computed by looking at the error in the different components of the formula. Now we claim to know $t$ to 0.01 second; but if you say that each of the two numbers we subtract is accurate to 0.01 second, then their difference will be inaccurate to 0.014 second (square root of 2 sneaks in when we take a difference). And that is using only the precision of the measurement, and assumes that everything else is in fact completely accurate (we have no other knowledge). Right there we have a 7% error in our $\Delta x$...
Why do we need the root 2? Basically, if they had measured the time difference to 1/1000000th of a second, and assuming that the clock used here only shows completely elapsed time (so it goes to 0.01 when the time is 0.01000 and not when it's 0.005), then you could have timed 148.159999 - 147.950000 = 0.209999 seconds, or 148.150000 - 147.959999 = 0.190001 seconds. In other words, it would follow a triangular distribution with a base of 0.02 s. Usually for error estimation, we assume the errors follow a normal distribution and estimate accordingly - you could get fancy and estimate the variance of the distribution, but that is overkill here (again I emphasize - we are estimating)
When we plug in the numbers, we get
$$\Delta x = \frac{1000}{148.05}{0.20} = 1.35 \pm 7\% = 1.35 \pm 0.10 m$$
Note that I used $148.05$ - the mean time - as the "nominal time taken". We don't know which track is correct, so we can't reliably compute the speed from either the slower or the faster time. This in turn gives us an error of $\frac{0.2}{148.05}\cdot100\%$ which is small compared to the error in $\Delta t$ at about 0.14% (compared to 7%).
It is not uncommon, when you have a good estimation of errors, to quote a result to a couple of additional significant figures and show the error estimate to multiple figures as well. In this case, our error estimate is not that good.
I believe it would be OK to quote the number as either $1.35 \pm 0.10 m$, or $1.4 \pm 0.1$.
Now as for the question "how much longer can the other track be"? That's a poorly phrased question - but according to my calculation above (of mean plus error), the track can be as long as 1.45 m (that would be within the error estimate). It then becomes a question of probabilities - I can say "with 50% confidence" that the track is 1.35 m longer, or with lower confidence I can state a greater difference.
Because of all these uncertainties, stating "1.4 m" is acceptable - although I would argue that people (especially physicists) need to be much better in stating their experimental error - both in the question, and in the answer. Having said that, there is an implied accuracy in a number - when you give a value like $1.4$, you imply the error is "on the order of" the least significant digit, i.e. that you mean $1.4 \pm 0.1m$ . The proposed answer with many significant figures, without an error estimate, is misleading.
Finally - it is usually good to reduce these things back to a conclusion that makes physical sense. A 1000 m course is usually run as 2.5 laps on a 400m course - so we might ask whether a distance in a measured course of $1.4/2.5m ~ 0.6m$ is reasonable. Since the track includes a complete circle, having the radius of the circle off by just 10 cm would be sufficient for this kind of error. I think that when people design athletic tracks very carefully, they try to do better than that - but it's really not a huge error.
Now if you had a 10 km course on a 400 m track, you are doing 25 laps - and the permissible error in the radius becomes even smaller. When you look at the fastest recorded times in the 10k (see for example http://en.wikipedia.org/wiki/10,000_metres), you will see that an inordinate number of these records were set at the same track - Brussels. I'm not claiming it's the length of the track - it might be the surface, the air quality, humidity, ... - but that's a LOT of fast runs (15 of the 25 fastest times) on just one venue. And spread out over a number of years. One can safely claim there is something "special" about that track... given that there are many years (recently - 1999, 2000, 2010, 2012, 2013) when nobody runs fast enough to be in this list, on any track in the world.
AFTERTHOUGHT
I re-read the question, and in particular the line
In order to conclude that the runner with the shorter time was indeed faster...
and am wondering whether this ought to be interpreted as a statistical hypothesis test. If you say that the null hypothesis is "they are going at the same speed, the tracks are different length" and the alternate hypothesis is "the track length alone cannot explain the difference in time", then you need to make a decision about the confidence level with which you reject the null hypothesis. So you need to ask yourself -
For a given velocities $v_1, v_2$ with error $\Delta v$, what is the track length difference for which $v_1 > v_2$ with > 95% confidence
That means we are looking at the error in velocity (which I estimated above at 0.07%) and choose the track lengths $T_1, T_2$ such that the velocity calculated from the times is different. This is a slightly different calculation than I had above, but it argues for rounding up the answer, rather than claiming that the mean difference is exact. To go from 50% to 95% confidence you need mean + 1.28 $\sigma$ (one sided confidence interval).
