a.
It is obvious that $1 \le \frac{2n^2 + 1}{n^2 - n + 1} \le 2$ for all $n \in \mathbb{Z}^+$. Hence, $0 \le a_n \le \ln 2 < 1$ and $\lfloor a_n \rfloor = 0$. It follows that $\{a_n\} = a_n$ and
$$\lim_{n \to \infty} \{a_n\} = \lim_{n \to \infty} a_n = \lim_{n \to \infty} \ln \frac{2n^2 + 1}{n^2 + n + 1} = \ln 2.$$
Therefore, there exists a positive integer $n_0$ such that
$$\{a_n\} > \ln 2 - \frac{1}{2016}, \forall n \ge n_0.$$
Then by direct calculation, it deduces that for $n > n_0$,
$$\{a_n\} = a_n > \ln 2 - \frac{1}{2016} > \frac{1}{2}.$$

b.
$$\lim_{n \to \infty} (b_n - b_{n-1}) = \lim_{n \to \infty} \ln \frac{(2n^2 + 1)(n^2 + n + 1)}{(2n^2 - 4n + 3)(n^2 - n + 1)} = 0.$$
Back to our problem, assume that there exists finite values of $n$ that $\{b_n\} < \frac{1}{2016}$. It follows that there exists a positive integer $n_0$ that $\{b_n\} \ge \frac{1}{2016}$ for all $n$. Because $\lim_{n \to \infty} (b_n - b_{n-1}) = 0$, it implies that there exists a positive integer $n_1$ that
$$b_n - b_{n-1} < \frac{1}{2016}$$
for all $n \ge n_1$. Because $(b_n)$ is an increasing sequence and $\lim_{n \to \infty} b_n = +\infty$, it follows that there exists infinite values of $n > \max\{n_0, n_1\}$ which $[b_n] - [b_{n-1}] = 1$. Consider such values of $n$, by the above inequality, we obtain
$$[b_n] - [b_{n-1}] + \{b_n\} - \{b_{n-1}\} < \frac{1}{2016}$$
or
$$\{b_{n-1}\} > \{b_n\} + \frac{2015}{2016}$$
Notice that $\{b_n\} \ge \frac{1}{2016}$, thus $\{b_{n-1}\} > 1$. This contradiction finishes the given problem. $\square$