a) For $a = 0$, we have $(u_n)$ defined by
$$u_1 = 6, u_{n+1} = 2 + \sqrt{u_n + 4}, \forall n \in \mathbb{N}^*.$$
It is clear that $u_n \ge 2$ for all positive integers $n$. On the other hand, we get $u_2 < u_1$. By induction, we can point out that $(u_n)$ is decreasing. Hence, $(u_n)$ has a finite limit $l$. By letting $n$ tend to infinity, we get $l = 2 + \sqrt{l + 4}$, thus $l = 5$.

b) Firstly, we prove that $(u_n)$ is bounded. Let $n_0$ be a positive integer such that $n_0 > a$. Now we choose a positive number $M > 10$ such that $M > \max(u_1, u_2, \dots, u_{n_0})$ which implies
$$u_{n_0+1} \le 3 + \sqrt{2u_{n_0} + 4} \le 3 + \sqrt{2M + 4} \le M.$$
By induction, we can point out that $u_n \le M$ for all positive integers $n$. Note that $u_n > 0$ then $(u_n)$ is bounded.
Next, we will prove this statement: $(u_n)$ is a monotone sequence (not necessary from the first term). Clearly, if $(u_n)$ is non-decreasing then the statement is proved. On the other hand, there exists a positive integer $m$ that $u_m > u_{m+1}$. Hence,
$$\frac{2m+a}{m} + \sqrt{\frac{m+a}{m} u_m + 4} > \frac{2m+2+a}{m+1} + \sqrt{\frac{m+1+a}{m+1} u_{m+1} + 4}.$$
Therefore, $u_{m+1} > u_{m+2}$. By induction, we claim that $(u_n)$ is decreasing from $u_m$ and the statement is proved. Note that we also have pointed out that $(u_n)$ is bounded, it implies that $(u_n)$ has a finite limit.