Is $\cup_{k=1}^\infty (r_k-\frac{1}{k}, r_k+\frac{1}{k}) = \mathbb{R}$? Let $r_k$ be the rational numbers in $\mathbb{R}$. 
(1).Is $\cup_{k=1}^\infty (r_k-\frac{1}{k^2}, r_k+\frac{1}{k^2}) = \mathbb{R}$?
(2).Is $\cup_{k=1}^\infty (r_k-\frac{1}{k}, r_k+\frac{1}{k}) = \mathbb{R}$?
(1).Because $m(\mathbb{R})=+\infty, \sum_{k=1}^\infty \frac{1}{k^2}<+\infty$, so $\mathbb{R} \setminus\cup_{k=1}^\infty (r_k-\frac{1}{k^2}, r_k+\frac{1}{k^2})\neq \Phi  $
(2) What about (2)?

I think (2) depends on how you enumerate the rationals. For example lets say you dont want $e$ in the image. Then enumerate the rationals so that if $q$ is a rational and 
$e-q \sim \frac{1}{n}$ the make sure that if $r_k=q$ we have $k > n$. 
(better is given in Ayman's answer and in the comments afterwards).
Conversely, there is an enumeration so that (2) is true, for example to cover the unit interval chose the rational sequence increasing so that you succsssively cover the interval, since the series diverges you will cover the whole interval, then do this over every interval.
I have given a very rough description of these constructions, they involve much back and forth to ensure one has a enumeration of all the rationls. 
