Let's consider the first inequality, and let's multiply both sides by $\frac{\sqrt{n}}2$: $$ \sqrt{n(n+1)}-n < \frac{1}{2} $$ Since $\lim_{n\to\infty}a_n=\frac{1}{2}$, it suffices to show that the sequence $a_n=\sqrt{n(n+1)}-n$ is strictly increasing, i.e. $a_n> a_{n-1}$ for $n>1$, so that $a_1< a_2 \ldots < a_n < \lim_{n\to\infty}a_n = \frac{1}{2}$ holds by induction. \begin{equation} a_n > a_{n-1} \quad\Leftrightarrow\quad \sqrt{n(n+1)}-n > \sqrt{n(n-1)}-n+1 \quad\Leftrightarrow\quad\\ n\left( \sqrt{1+\frac{1}{n}} - \sqrt{1-\frac{1}{n}} \right) > 1 \end{equation} Let's raise both sides to the second power: $$ \quad\Leftrightarrow\quad 2n^2\left( 1 - \sqrt{1-\frac{1}{n^2}} \right) > 1 \quad\Leftrightarrow\quad 2n^2 - 1 > 2n\sqrt{n^2-1} \quad\Leftrightarrow\quad\\ 4n^4 + 1 - 4n^2 > 4n^4 - 4n^2 \quad\Leftrightarrow\quad 1>0 $$ so the first inequality is proven.
Concerning the second one, for the same reasoning you show that the sequence $b_n=n-\sqrt{n(n-1)}$ is strictly decreasing (for simpler calculations, I'd suggest to show that $b_{n+1}>b_n$ rather than $b_n> b_{n-1}$); actually you end up with showing that $\sqrt{n(n+1)}-\sqrt{n(n-1)}>1$, which we proved above.