The continued fraction representations of the limits of the interval are
$$ 0.0080120180265 = [0; 124, 1, 4, 2, 1, 463872, 1, 1, 12, 1, 1, 41] \\
0.0080120180275 = [0; 124, 1, 4, 3, 545777, 2, 13, 1, 1, 1, 1, 2] $$
The simplest continued fraction (and therefore also the simplest ordinary fraction!) in that interval is
$$ [0;124,1,4,3] = \frac{16}{1997} = 0.00801201802704056084\ldots $$
and the sum of its numerator and denominator is $2013$.
(I used Wolfram Alpha to expand the continued fractions fully. For a pencil-and-paper solution one only needs to carry out the expansion until they start differing, which requires only a handful of long divisions with remainder.)
One has $8!=2^7\cdot 3^2\cdot 5\cdot 7$ and ${\root3\of 8!}\approx34.3$. We now should factor $8!$ into three factors as equal as possible, which means: as near to $34.3$ as possible.. It seems that $a=2^5=32$, $b=5\cdot 7=35$, $c=2^2\cdot3^2=36$ is the best we can do. Note that neither $33$ nor $34$ can be attained with the primes at disposal. The minimal possible value of $c-a$ therefore is $4$.
Best Answer
AM-GM inequality is a good idea: \begin{align} & (a + b + c)\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c}\right) \\ = & \frac{1}{2}((a + b) + (a + c) + (b + c))\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c}\right) \\ \geq & \frac{1}{2} \times 3\sqrt[3]{(a + b)(a + c)(b + c)} \times 3\sqrt[3]{\frac{1}{(a + b)(a + c)(b + c)}} \\ = & \frac{9}{2} \end{align}
The equality holds when $a + b = a + c = b + c$.