Chọn phương án D.

\(\begin{aligned}
2^{x-2}=3^{x^2+2x-8}\Leftrightarrow&2^{x-2}=3^{(x+4)(x-2)}\\
\Leftrightarrow&2^{x-2}=\left(3^{x+4}\right)^{x-2}\\
\Leftrightarrow&1=\left(\dfrac{3^{x+4}}{2}\right)^{x-2}\\
\Leftrightarrow&\left[\begin{array}{l}
x-2=0\\
\dfrac{3^{x+4}}{2}=1
\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}
x=2\\
3^{x+4}=2
\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}
x=2\\
x+4=\log_32
\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}
x=2\\
x=\log_32-4.
\end{array}\right.
\end{aligned}\)

Vậy \(a=3\) và \(b=2\).

Suy ra \(a+2b=3+2\cdot2=7\).