Are there problems with complexity between $O(n^c/\log^b n)$ and $O(n^{c - \varepsilon})$?
The question is about upper bound (which is $O(n^{2013} / (\log n)^{2012})$) and lower bound (which is $O(n^{2012.9})$) &ndash; do they correspond to each other?  

According to definitions of Big-O and &Omega; we have to find constants $c_1$, $c_2$, and $n_0$ such that for $n\geq n_0$ two following inequalities are true:

$$c_1 n^{2012.9} \leq T(n) \leq c_2 n^{2013} / (\log n)^{2012}.$$

If such constants exist, then the lower bound must be less than upper bound for sufficiently large $n$:

$$c_1 n^{2012.9} \leq c_2 n^{2013} / (\log n)^{2012}.$$

Let's simplify this a little:

$$c_1 (\log n)^{2012} \leq c_2 n^{0.1}.$$

Can we find $n_0$ such that this inequality is true when $n \geq n_0$? Yes, sure &ndash; the power function eventually wins over the logarithm function, so the answer is TRUE.