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The function $f(x)=\frac{2-\sqrt{x+4}}{\sin 2 x}, x \neq 0$, is continuous at $x=0$.

$\therefore f(0)$

$=\lim _{x \rightarrow 0} f(x)$

$=\lim _{x \rightarrow 0} \frac{2-\sqrt{x+4}}{\sin 2 x}$

$=\lim _{x \rightarrow 0} \frac{2-\sqrt{x+4}}{\sin 2 x} \times \frac{2+\sqrt{x+4}}{2+\sqrt{x+4}}$

$=\lim _{x \rightarrow 0} \frac{4-(x+4)}{\sin 2 x(2+\sqrt{x+4})}$

$=\lim _{x \rightarrow 0} \frac{-x}{\sin 2 x(2+\sqrt{x+4})}$

$=-\frac{1}{2} \times \lim _{x \rightarrow 0} \frac{2 x}{\sin 2 x(2+\sqrt{x+4})}$

$=-\frac{1}{2} \times \frac{1}{\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{2 x}\right)} \times \frac{1}{\lim _{x \rightarrow 0}(2+\sqrt{x+4})}$

$=-\frac{1}{2} \times \frac{1}{1} \times \frac{1}{(2+\sqrt{0+4})} \quad\left(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right)$

$=-\frac{1}{2} \times \frac{1}{4}$

$=-\frac{1}{8}$

Thus, the value of $f(0)$ is $-\frac{1}{8}$.

If $f(x)=\frac{2-\sqrt{x+4}}{\sin 2 x}, x \neq 0$, is continuous at $x=0$, then $f(0)=-\frac{1}{8}$