1. Two quadratic equations with unequal leading coefficients
$$
\begin{array}{l}
(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0, \\
(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 b\right)=0
\end{array}
$$
$\left(a 、 b \in \mathbf{N}_{+}\right)$ have a common root. Find the value of $\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}$.
Given the known equations, we have $a \neq 1, b \neq 1$. Therefore, $a > 1, b > 1$, and $a \neq b$. Let $x_{0}$ be the common root of the two equations. It is easy to see that $x_{0} \neq 1$. By the definition of the roots of the equation, $a$ and $b$ are the two distinct real roots of the equation
$$
\left(1-x_{0}\right) t^{2}+\left(2+x_{0}^{2}\right) t-\left(2 x_{0}+x_{0}^{2}\right)=0
$$

Then,
$$
\begin{array}{l}
a+b=\frac{2+x_{0}^{2}}{x_{0}-1}, \quad a b=\frac{2 x_{0}+x_{0}^{2}}{x_{0}-1} \\
\Rightarrow (a-1)(b-1)=3 \\
\Rightarrow \{a, b\}=\{2,4\} \\
\Rightarrow \frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}=a^{b} b^{a}=2^{4} \times 4^{2}=256 .
\end{array}
$$