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.)
1.$S=\frac{t_1}{1-t_2}=\frac{1/r}{1-r}=\frac{1}{r-r^2}$. Finding min for $S$ is the same as max for $\frac{1}{S}=r-r^2$, where the max is at $r=\frac{1}{2}$, leading to $S=4$ as the minimum.
- $20180=(2018)^2\frac{(1-r)^2}{1-r}$ So $10=2018(1-r)$ or $r=\frac{2008}{2018}=\frac{1004}{1009}$ where the numerator and denominator are relatively prime. So $m+n=2013$.
Best Answer
The way to find the smallest denominator "from scratch" is with continued fractions.
Begin by rendering the proposed bounds thusly:
$\dfrac{7}{10}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{3}}}}$
$\dfrac{11}{15}=\dfrac{1 }{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{1+\dfrac{1}{3}}}}}$
The upper layers of the continued fractions are identical but they eventually become different when we get down to the layers in blue. We may now replace those entries with the smallest whole number lying between, thus
$1+\dfrac{1}{3}<2<3$
So the smallest denominator fraction meeting the betweenness criterion will be
$\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{2}}}}=\dfrac{5}{\color{blue}{7}}$