Lebesgue Measure - positive measure sets not containing intervals

For 1: There is an open set $U$ containing all rationals such that $m(U)< (b-a).$ Thus $m([a,b]\setminus U) > 0.$ The set $[a,b]\setminus U$ is compact, is a subset of $[a,b],$ and since it contains no rational, has no interior.

My reaction is just to trace through some examples. The set $K$ in part $1$ is something like a Cantor set-just fattened up to maintain positive measure, say for concreteness $1/2$. While we're simplifying, may as well take $[a,b]=[0,1]$ as well. $K$ needn't be uniformly distributed-indeed the point of your part 2 is that it can't be-so we may suppose we can remove some interval and increase the proportional measure of $K$. In the most usual construction, we might have $m(K\cap [0,3/8])=1/2$. Then if $K$ is self-symmetrical, as it may certainly be, we immediately get arbitrarily large proportions of $K$ in sufficiently small intervals.

This is all in contrast to the case of the usual Cantor set $C$: one way to prove $m(C)=0$ is to observe that for every interval $I$, $C$ is contained in a subset of $I$ of relative measure no more than $2/3$. For the construction of $C$ is uniform: at every stage we remove intervals of relative measure $2/3$. In contrast again this shows why we don't construct $K$ by uniform removal of even very small intervals.