Answer
$$
\lim _{x \rightarrow 0} \frac{\sin (a+2 x)-2 \sin (a+x)+\sin a}{x^{2}}=-\sin a
$$
Work Step by Step
$$
\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin (a+2 x)-2 \sin (a+x)+\sin a}{x^{2}}= \\=& \lim _{x \rightarrow 0} \frac{\sin a \cos 2 x+\cos a \sin 2 x-2 \sin a \cos x-2 \cos a \sin x+\sin a}{x^{2}}= \\=& \lim _{x \rightarrow 0} \frac{\sin a(\cos 2 x-2 \cos x+1)+\cos a(\sin 2 x-2 \sin x)}{x^{2}} =\\=& \lim _{x \rightarrow 0} \frac{\sin a\left(2 \cos ^{2} x-1-2 \cos x+1\right)+\cos a(2 \sin x \cos x-2 \sin x)}{x^{2}} =\\
&=\lim _{x \rightarrow 0} \frac{\sin a(2 \cos x)(\cos x-1)+\cos a(2 \sin x)(\cos x-1)}{x^{2}}= \\
&
=\lim _{x \rightarrow 0} \frac{2(\cos x-1)[\sin a \cos x+\cos a \sin x](\cos x+1)}{x^{2}(\cos x+1)} =\\
& =\lim _{x \rightarrow 0} \frac{-2 \sin ^{2} x[\sin (a+x)]}{x^{2}(\cos x+1)}= \\ &=-2 \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{2} \cdot \frac{\sin (a+x)}{\cos x+1}=\\
& =-2(1)^{2} \frac{\sin (a+0)}{\cos 0+1}=\\
&=-\sin a
\end{aligned}
$$
So ,
$$
\lim _{x \rightarrow 0} \frac{\sin (a+2 x)-2 \sin (a+x)+\sin a}{x^{2}}=-\sin a
$$