For part a) we easy have $u_n>5\quad\forall n\ge 1$ and $$|u_{n+1}-5|=\frac{|u_n-5|}{\sqrt{u_n+4}+3}<\frac{1}{5}|u_n-5|<\cdots<\left(\frac{1}{5}\right)^n|u_1-5|\implies \lim u_n=5.$$For part b) we use the well-known lemma If $z_n$ is positive sequence and $z_{n+1}\le cz_n+q_n$ with $\lim q_n=0$ and $c\in(0,1)$ then we have $\lim z_n=0$. Let $v_n=\frac{u_n}{n}$ then we have $v_1=6$ and $$v_{n+1}=\frac{\frac{2n+a}{n}+\sqrt{(n+a)v_n+4}}{n+1}<\frac{2n+a}{n^2+n}+\frac{\sqrt{(n+a)v_n}}{n+1}+\frac{2}{n+1}\\ \left(\sqrt{a+b}<\sqrt{a}+\sqrt{b}\right)$$$$\overset{\text{AM-GM}}{\le} \frac{2n+a}{n^2+n}+\frac{2}{n+1}+\frac{1}{2}\left(\frac{1}{n+1}+\frac{n+a}{n+1}v_n\right)=\frac{2n+a}{n^2+n}+\frac{\frac{5}{2}}{n+1}+\frac{1}{2}\cdot\frac{n+a}{n+1}v_n$$Since $\lim \frac{n+a}{n+1}=1$ so there exist $N$ such that forall $n\ge N$ one has $\frac{n+a}{n+1}<\frac{3}{2}$. So for $n\ge N$ we have $$v_{n+1}<\left(\frac{2n+a}{n^2+n}+\frac{5}{2(n+1)}\right)+\frac{3}{4}v_n$$By the lemma hence $\lim \frac{u_n}{n}=0$. We have $$|u_{n+1}-5|=\left|\frac{u_n-5+\frac{au_n}{n}}{\sqrt{\frac{n+a}{n}.u_n+4}+3}+\frac{a}{n}\right|<\frac{1}{5}|u_n-5|+\left(\frac{a}{5}\cdot\frac{u_n}{n}+\frac{a}{n}\right)\ (u_n>5)$$By $\lim\frac{u_n}{n}=0$ and $\frac{a}{n}=0$ applying the lemma we have $\lim |u_n-5|=0$ or $u_n$ has limit $5$

DNCT1. You can prove that lemma: If $z_n$ is positive sequence and $z_{n+1}\le cz_n+q_n$ with $\lim q_n=0$ and $c\in(0,1)$ then we have $\lim z_n=0$.