(7) Let $\left(x^{2}+2 x-2\right)^{6}=a_{0}+a_{1}(x+2)+a_{2}(x+2)^{2}+\cdots+a_{12}(x+$ $2)^{12}$, where $a_{i}(i=0,1,2, \cdots, 12)$ are real constants, then $a_{0}+a_{1}+2 a_{2}+3 a_{3}+\cdots+$ $12 a_{12}=$ $\qquad$ .
(7) By setting \( x = -2 \) in the given equation, we obtain \( a_{0} = 64 \). Differentiating both sides of the given equation,

we get
$$
\begin{aligned}
& 6\left(x^{2}+2 x-2\right)^{5} \cdot (2 x+2) \\
= & a_{1} + 2 a_{2}(x+2) + 3 a_{3}(x+2)^{2} + \cdots + 12 a_{12}(x+2)^{11},
\end{aligned}
$$

Substituting \( x = -1 \), we get
$$
a_{1} + 2 a_{2} + 3 a_{3} + \cdots + 12 a_{12} = 0.
$$

Therefore,
$$
a_{0} + a_{1} + 2 a_{2} + 3 a_{3} + \cdots + 12 a_{12} = 64 + 0 = 64.
$$