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For general Diophantine approximation, the Thue–Siegel–Roth theorem states that for any irrational algebraic number $x$, and any $\varepsilon>0$, there exists a constant $c=c(x,\varepsilon)$ such that $$ \left|x-\frac{p}q\right| > \frac{c}{q^{2+\varepsilon}}, $$ for any integers $p$ and $q>0$.

My first question is, is it possible to get larger lower bound, if we are allowed to choose $x$ carefully?

The second question is, does there exist $x$ such that $$ \left|x-\frac{p}{2^k}\right| > \frac{c}{2^{k}}, $$ for any $p$ and $k$, with some constant $c>0$?

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$x=1/3$? ${}{}{}$ –  Yoav Kallus Jul 5 '13 at 21:20
@YoavKallus: Thanks! Does it work? Sorry if the question is too easy. Perhaps this should be moved to MSE? –  timur Jul 5 '13 at 22:49
The binary expansion is $1/3=0.010101010101\ldots$, so for any fixed $k$, $\min_p 2^k |x-p2^{-k}| = 0.010101\ldots = 1/3$. So that should answer your second question. As for the first, I'm not sure, but probably the golden ratio should work if any number does. –  Yoav Kallus Jul 5 '13 at 23:11
Indeed, this question indicates it does work for the golden ratio, as for any number whose continued fraction expansion is bounded: mathoverflow.net/questions/28347/… –  Yoav Kallus Jul 5 '13 at 23:19
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So if you pick $x$ by specifying the continued fraction, and it is bounded, then you will have $\left| x-\frac{p}{q} \right| \gt \frac{c}{q^{2}}$ However it is unlikely (with random choice) to be algebraic. The continued fraction expansion of a number $x$ is eventually priodic (and hence, of course, bounded) exactly if $x$ is of the form $\frac{u+\sqrt{v}}{w}$ (i.e. is algebraic of degree $2$.) I do not know if any algebraic number of degree $3$ or greater is known to have a bounded or known to have an unbounded continued fraction, however there are reasons to think that they are all unbounded.

The binary expansion of $x$ plays a similar role to your second question. Any real number who binary expansion has a bounded length to the runs of consecutive identical bits (and only such real numbers) will satisfy a bound $\left|x-\frac{p}{2^k}\right| > \frac{c}{2^{k}}$ where $c$ depends on the longest run of $0$'s or $1$'s . This is automatic if the binary expansion is eventually periodic, which means that $x$ is rational (and not of the form $\frac{n}{2^k}$ .) Many other examples can be given by specifying an appropriate binary expansion, but perhaps not in any other (known) way.

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