2. If $(2 x+4)^{2 n}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n}\left(n \in \mathbf{N}^{+}\right)$, then the remainder when $a_{2}+a_{4}+\cdots+a_{2 n}$ is divided by 3 is . $\qquad$
$\begin{array}{l}\text { 2. } 1 \text { Detailed Explanation: } a_{0}+a_{2}+a_{4}+\cdots+a_{2 n}=\frac{1}{2}\left[(2+4)^{2 n}+(-2+4)^{2 n}\right]=\frac{1}{2}\left[6^{2 n}+2^{2 n}\right] \text {. } \\ a_{2}+a_{4}+\cdots+a_{2 n}=2^{2 n-1}\left(3^{2 n}+1\right)-4^{2 n} \equiv(-1)^{2 n-1} \times 1-1^{2 n}=-2 \equiv 1(\bmod 3) .\end{array}$