Uniform convergence over $[0, \alpha]$. I presume you are trying the apply the inequality to bound the given expression directly. Though this is correct, this is not useful because the inequality $\sin \theta \geqslant \theta - \frac{\theta^3}{3!}$ ($\theta \geqslant 0$) is tight only for small $\theta$, whereas the expression inside the "sin" grows unbounded in our case. So we massage the function a little before employing the inequality.
$$
\begin{eqnarray*}
\sin (\sqrt{4\pi^2 n^2 + x^2})
&=&
\sin (\sqrt{4\pi^2 n^2 + x^2} - 2 n \pi)
\\ &=&
\sin \left(\frac{4\pi^2 n^2 + x^2 - (2 n \pi)^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} \right)
\\ &=&
\sin \left(\frac{x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} \right).
\end{eqnarray*}
$$
Notice that $\frac{x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi}$ is small, so we can hope to apply the inequality at this stage. Doing so gives
$$
\begin{eqnarray*}
\frac{x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi}
&\geqslant& \sin (\sqrt{4\pi^2 n^2 + x^2})
\\ &\geqslant& \frac{x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} - \frac{1}{3!} \left( \frac{x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} \right)^3.
\end{eqnarray*}
$$
Try to take it from here.
EDIT: More steps added. The left hand side inequality implies that
$$
\frac{x^2}{4\pi} - n\sin(\sqrt{4 \pi^2 n^2 + x^2}) \geqslant 0,
$$
so we only need to upper bound the difference. For this, we use the right hand side inequality:
$$
\begin{eqnarray*}
\frac{x^2}{4 \pi} - n\sin(\sqrt{4 \pi^2 n^2 + x^2})
&\leqslant& \frac{x^2}{4 \pi} - \frac{n x^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} + \frac{n x^6}{6 (\color{Green}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi})^3}
\\ &\leqslant& \frac{x^2}{4 \pi} - \frac{nx^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi} + \frac{nx^6}{6 (\color{Green}{4n\pi})^3}
\\ &\leqslant& \color{Red}{\frac{x^2}{4 \pi}} \left[ 1 - \frac{2}{\sqrt{1 + \frac{x^2}{4 \pi^2n^2}} + 1} \right] + \color{Red}{\frac{nx^6}{6 (4n\pi)^3}}
\\ &\leqslant& \color{Red}{\frac{{\alpha}^2}{4 \pi}} \left[ \color{Blue}{ 1 - \frac{2}{\sqrt{1 + \frac{x^2}{4 \pi^2n^2}} + 1} } \right] + \color{Red}{\frac{n{\alpha}^6}{6 (4n\pi)^3}}
\\ &\leqslant& \frac{\alpha^2}{4 \pi} \left[ \color{Blue}{ 1 - \frac{2}{\sqrt{1 + \frac{\alpha^2}{4 \pi^2n^2}} + 1}} \right] + \frac{n\alpha^6}{6 (4n\pi)^3},
\end{eqnarray*}
$$
repeatedly using the fact that $0 \leqslant x \leqslant \alpha$.
Non-uniform convergence over $[0, \infty)$. We saw that
$$
\begin{eqnarray*}
n\sin (\sqrt{4\pi^2 n^2 + x^2})
&\leqslant& \frac{nx^2}{\sqrt{4\pi^2 n^2 + x^2} + 2 n \pi}
\\ &=& \frac{x^2}{4 \pi} \frac{2}{\sqrt{1 + \frac{x^2}{4\pi^2n^2}} + 1} .
\end{eqnarray*}
$$
Can you get the inequality claimed by the hint from here?
EDIT: More hints on the non-uniform convergence. This problem is meant to highlight the (subtle at first glance) difference between pointwise and uniform convergence. For any fixed $x$, as $n \to \infty$, it is true that the sequence $f_n(x)$ converges to $f(x)$; but this is what pointwise convergence is about.
If we want to show uniform convergence, we need to show that the sequence
$$
u_n := \sup \{ |f_n(x) - f(x)| \colon x \in \mathbb R \}
$$
goes to $0$ as $n \to \infty$. The idea is to provide a uniform upper bound on the error term $|f_n(x) - f(x)|$ that is independent of $x \in \mathbb R$, such that the upper bound goes to $0$ as $n \to \infty$.
In our example, we are interested in the sequence
$$
\left| n\sin (\sqrt{4\pi^2 n^2 + x^2}) - \frac{x^2}{4 \pi} \right| = \frac{x^2}{4 \pi} - n\sin (\sqrt{4\pi^2 n^2 + x^2}) .
$$
To prove non-uniform convergence (over $[0, \infty)$), we need to lower bound the sequence
$$
u_n := \sup_{x \in [0, \infty)} \left( \frac{x^2}{4 \pi} - n\sin (\sqrt{4\pi^2 n^2 + x^2}) \right) .
$$
By the inequality given to you in the text-book hint, we have
$$
\begin{eqnarray*}
u_n
&\geqslant&
\sup_{x \in [0, \infty)} \frac{x^2}{4\pi} \left ( 1-\frac{2}{\sqrt{1+\frac{x^2}{4\pi^2n^2}}+1} \right) .
\\ &\geqslant&
\sup_{x \in [0, \infty)} \frac{x^2}{4\pi} \left ( 1-\frac{2}{\Big(\frac{|x|}{2 n \pi} \Big)} \right) .
\\ &=&
\sup_{x \in [0, \infty)} \left( \frac{x^2}{4\pi} - n|x| \right) .
\end{eqnarray*}
$$
Taking $x = 8 n \pi$, we get
$$
\begin{eqnarray*}
u_n
\geqslant
\frac{64 n^2 \pi^2}{4 \pi} - n \cdot 8 n \pi = 8 n^2 \pi \to \infty,
\end{eqnarray*}
$$
as $n \to \infty$. Therefore, clearly, $u_n$ cannot approach $0$ as $n \to \infty$. In fact, by choosing $x$ a little more carefully,* you can show that $u_n=\infty$ for each $n$.
*Note: Notice that our choice of $x$ varies as $n$ varies. This is unavoidable because we already know that for any fixed $x$, the above deviation term goes to $0$.