Chọn phương án A.

\begin{eqnarray*}
&y&=\sin x-\cos x+3\\
\Leftrightarrow&\dfrac{y}{\sqrt{2}}&=\dfrac{1}{2}\sin x-\dfrac{1}{2}\cos x+\dfrac{3}{\sqrt{2}}\\
\Leftrightarrow&\dfrac{y}{\sqrt{2}}&=\cos\dfrac{\pi}{3}\sin x-\sin\dfrac{\pi}{3}\cos x+\dfrac{3}{\sqrt{2}}\\
\Leftrightarrow&\dfrac{y}{\sqrt{2}}&=\sin\left(x-\dfrac{\pi}{4}\right)+\dfrac{3}{\sqrt{2}}\\
\Leftrightarrow&y&=\sqrt{2}\sin\left(x-\dfrac{\pi}{4}\right)+3.
\end{eqnarray*}
Mặt khác,
\begin{eqnarray*}
-1\le&\sin\left(x-\dfrac{\pi}{4}\right)&\le1\\
\Leftrightarrow-\sqrt{2}\le&\sqrt{2}\sin\left(x-\dfrac{\pi}{4}\right)&\le\sqrt{2}\\
\Leftrightarrow3-\sqrt{2}\le&\sqrt{2}\sin\left(x-\dfrac{\pi}{4}\right)+3&\le3+\sqrt{2}\\
\Leftrightarrow3-\sqrt{2}\le&y&\le3+\sqrt{2}
\end{eqnarray*}
Theo đó, $\begin{cases}
M=3+\sqrt{2}\\
m=3-\sqrt{2}
\end{cases}$. Vậy $M\cdot m=7$.