a) We shall prove the statement by induction if $a_1 > \frac{19}{243} \left(\frac{1}{12} > \frac{19}{243}\right)$. It is true for $n \le 3$, since $a_2 > \sqrt{3 \cdot \frac{19}{243} + 1} = \frac{10}{9}$ and $a_3 > \sqrt{4 \cdot \frac{10}{9} + 1} = \frac{7}{3}$.

Assume that $a_n > n - \frac{2}{n}$ for some $n \ge 3$. Then
$$a_{n+1} > \sqrt{(n+2)\left(n - \frac{2}{n}\right) + 1}.$$
It is enough to show that right-hand side is bigger than $n+1 - \frac{2}{n+1}$. It is easy to see that this is equivalent to $\frac{1}{2} > \frac{1}{n} + \frac{1}{(n+1)^2}$, which obviously holds for $n \ge 3$.

b) Note first that if $a_1 = 1$, then $a_n = n$ by induction, implying that $b_n = 2^n \left(\frac{a_n}{n} - 1\right) = 0$.
If $a_1 < 1$, then $a_n < n$ again by induction, i.e., $b_n < 0$. We shall prove that $b_n < b_{n+1}$. This is equivalent to $\frac{a_n - n}{2n} < \frac{a_{n+1} - n - 1}{n+1}$. The right-hand side equals
$$\frac{a_{n+1}^2 - n - 1}{(n+1)(a_{n+1} + n + 1)} = \frac{(n+2)a_n + 1 - (n+1)^2}{(n+1)(a_{n+1} + n + 1)} = \frac{(n+2)(a_n - n)}{(n+1)(a_{n+1} + n + 1)}.$$
It remains to show that $\frac{1}{2n} > \frac{n+2}{(n+1)(a_{n+1} + n + 1)}$, i.e.,
$$(n+1)a_{n+1} > 2n(n+2) - (n+1)^2 = (n+1)^2 - 2,$$
which follows by a).
The above arguments show that if $a_1 > 1$, then $b_n > b_{n+1} > 0$.
So the sequence $(b_n)$ is monotone and bounded; hence it converges.