We will solve directly part 2), from which to deduce the result of part 1). It can be seen that the sequence $(u_n)$ defines if and only if $u_2$ defines. Since $u_2 = \frac{1}{2} + \sqrt{\frac{5}{2}a + \frac{1}{4}}$, then $u_2$ defines if and only if
$$a \geq -\frac{1}{10}.$$
We will prove that the sequence $(u_n)$ converges to 3 for every $a \geq -\frac{1}{10}$. It is easy to see that $u_n \geq \frac{1}{2}, \forall n \geq 2$. Note that $f(x) = \frac{2x+3}{x+1}$ is a strictly decreasing function over $\mathbb{R}^+$ so for every positive integers $n$, we have
$$\frac{2n+3}{n+1} > \frac{2(n+1)+3}{(n+1)+1}.$$
If there exists a number $n_0 \in \mathbb{N}$ such that $u_{n_0} \ge u_{n_0+1}$ then we have
$$u_{n_0+2} = \frac{1}{2} + \sqrt{\frac{2(n_0+1)+3}{(n_0+1)+1}u_{n_0+1} + \frac{1}{4}} \le \frac{1}{2} + \sqrt{\frac{2n_0+3}{n_0+1}u_{n_0} + \frac{1}{4}} = u_{n_0+1}.$$
Similarly, we have $u_{n_0} \ge u_{n_0+1} \ge u_{n_0+2} \ge u_{n_0+3} \ge \dots$. This means that the sequence $(u_n)$ does not increase from $n_0$ onward. Simultaneously, the sequence is also bounded lower for every $n \ge 2$ so there exists a limit $\lim_{n \to \infty} u_n = L$ ($L \ge \frac{1}{2}$). Now, convert equation (1) to limit, we get
$$L = \frac{1}{2} + \sqrt{2L + \frac{1}{4}}.$$
Solve this equation, we have $L=3$. In summary, if there exists $n_0$ as above then $\lim_{n \to \infty} u_n = 3$.

Next, we consider the case: the number $n_0$ as above does not exist, in other words the sequence $(u_n)$ is strictly decreasing. We will prove $(u_n)$ is also bounded above. Indeed, since $u_{n+1} > u_n$ and $\frac{2n+3}{n+1} < 3$ for every $n \ge 2$ so we have
$$\frac{1}{2} + \sqrt{3u_n + \frac{1}{4}} > \frac{1}{2} + \sqrt{\frac{2n+3}{n+1}u_n + \frac{1}{4}} = u_{n+1} > u_n, \quad \forall n \ge 2.$$
Solving this inequality of $u_n$, we have $u_n < \frac{4+\sqrt{17}}{2}$ for every $n \ge 2$. Therefore the sequence $(u_n)$ is strictly decreasing and bounded upper by $\frac{4+\sqrt{17}}{2}$ so it has a finite limit. Hence, by convert equation (1) to limit, we also have $\lim_{n \to \infty} u_n = 3$.

In summary, for every $a \ge -\frac{1}{10}$ the sequence $(u_n)$ defines and converges to 3. ■