Chọn phương án C.

Với $0<x\neq1$, đặt $t=\log_2x$ ta có $$\begin{array}{lll}
&2\log_2x+3\log_x2&=7\\
\Leftrightarrow&2\log_2x+\dfrac{3}{\log_2x}&=7\\
\Leftrightarrow&2t+\dfrac{3}{t}&=7\\
\Leftrightarrow&2t^2-7t+3&=0\\
\Leftrightarrow&\left[\begin{array}{l}t=3\\ t=\dfrac{1}{2}\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}\log_2x=3\\ \log_2x=\dfrac{1}{2}\end{array}\right.\Leftrightarrow\left[\begin{array}{l}x=8\\ x=\sqrt{2}.\end{array}\right.
\end{array}$$
Theo đề bài thì $x_1=\sqrt{2}$, $x_2=8$.
Vậy $T=\big(x_1\big)^{x_2}=\sqrt{2}^8=16$.