On tap tonight...

If the two taps chosen can fill the tank in $m$ and $n$ hours, respectively, then together they can fill the tank at a rate of

$\dfrac{1}{m} + \dfrac{1}{n} = \dfrac{m + n}{mn}$ "tanks" per hour.

The time taken to fill the tank is then the reciprocal of this expression, namely $\dfrac{mn}{m + n}$ hours.

Now there are $\binom{5}{2} = 10$ possible combinations of $2$ taps chosen from the $5$ available. Since each of these combinations is equally likely, to find the expected time to fill the tank we just need to sum the $10$ possible times and then divide the result by $10$ . This gives us an expected time of

$\frac{1}{10}*(\frac{2}{3} + \frac{3}{4} + \frac{4}{5} + \frac{5}{6} + \frac{6}{5} + \frac{8}{6} + \frac{10}{7} + \frac{12}{7} + \frac{15}{8} + \frac{20}{9}) = \dfrac{6463}{5040}$ .

Thus $a = 6463, b = 5040$ and $a - b = \boxed{1423}$ .