答案：
(1)【证】由已知得
$$
\boldsymbol{A} \boldsymbol{\beta}=\boldsymbol{A}\left(\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2+\boldsymbol{\alpha}_3\right)=\lambda_1 \boldsymbol{\alpha}_1+\lambda_2 \boldsymbol{\alpha}_2+\lambda_3 \boldsymbol{\alpha}_3 .
$$
若 $\boldsymbol{\beta}$ 是 $\boldsymbol{A}$ 的特征向量, 假设对应的特征值为 $\mu$, 则
$$
\begin{aligned}
& \boldsymbol{A} \boldsymbol{\beta}=\boldsymbol{\beta}=\mu\left(\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2+\boldsymbol{\alpha}_3\right) \\
\Rightarrow & \lambda_1 \boldsymbol{\alpha}_1+\lambda_2 \boldsymbol{\alpha}_2+\lambda_3 \boldsymbol{\alpha}_3=\mu\left(\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2+\boldsymbol{\alpha}_3\right) \\
\Rightarrow & \left(\mu-\lambda_1\right) \boldsymbol{\alpha}_1+\left(\mu-\lambda_2\right) \boldsymbol{\alpha}_2+\left(\mu-\lambda_3\right) \boldsymbol{\alpha}_3=\mathbf{0} .
\end{aligned}
$$
由于 $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3$ 是不同特征值对应的特征向量, 故 $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3$ 线性无关, 则 $\lambda_1=\lambda_2=\lambda_3=\mu$, 与 $\lambda_1, \lambda_2$, $\lambda_3$ 是 $\boldsymbol{A}$ 的 3 个不同特征值矛盾,故 $\boldsymbol{\beta}$ 不是 $\boldsymbol{A}$ 的特征向量.
(2)【证】法一(用定义) 设 $k_1 \boldsymbol{\beta}+k_2 \boldsymbol{A} \boldsymbol{\beta}+k_3 \boldsymbol{A}^2 \boldsymbol{\beta}=\mathbf{0}$, 代人 $\boldsymbol{\beta}=\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2+\boldsymbol{\alpha}_3$, 并注意 $\boldsymbol{A}^n \boldsymbol{\alpha}=\lambda^n \boldsymbol{\alpha}$, 有
$$
\begin{aligned}
& k_1\left(\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2+\boldsymbol{\alpha}_3\right)+k_2\left(\lambda_1 \boldsymbol{\alpha}_1+\lambda_2 \boldsymbol{\alpha}_2+\lambda_3 \boldsymbol{\alpha}_3\right)+k_3\left(\lambda_1^2 \boldsymbol{\alpha}_1+\lambda_2^2 \boldsymbol{\alpha}_2+\lambda_3^2 \boldsymbol{\alpha}_3\right)=\mathbf{0} \\
\Rightarrow & \left(k_1+k_2 \lambda_1+k_3 \lambda_1^2\right) \boldsymbol{\alpha}_1+\left(k_1+k_2 \lambda_2+k_3 \lambda_2^2\right) \boldsymbol{\alpha}_2+\left(k_1+k_2 \lambda_3+k_3 \lambda_3^2\right) \boldsymbol{\alpha}_3=\mathbf{0} .
\end{aligned}
$$
由于 $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3$ 是不同特征值 $\lambda_1, \lambda_2, \lambda_3$ 对应的特征向量,故 $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3$ 线性无关. 则
$$
\left\{\begin{array}{l}
k_1+\lambda_1 k_2+\lambda_1^2 k_3=0 \\
k_1+\lambda_2 k_2+\lambda_2^2 k_3=0 \\
k_1+\lambda_3 k_2+\lambda_3^2 k_3=0
\end{array}\right.
$$
由 $\left|\begin{array}{lll}1 & \lambda_1 & \lambda_1^2 \\ 1 & \lambda_2 & \lambda_2^2 \\ 1 & \lambda_3 & \lambda_3^2\end{array}\right| \neq 0$, 故方程组只有零解, 即 $k_1=k_2=k_3=0$, 所以 $\boldsymbol{\beta}, \boldsymbol{A} \boldsymbol{\beta}, \boldsymbol{A}^2 \boldsymbol{\beta}$ 线性无关.
法二(用秩) 由 $\boldsymbol{\beta}=\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2+\boldsymbol{\alpha}_3$, 并注意 $\boldsymbol{A}^n \boldsymbol{\alpha}=\lambda^n \boldsymbol{\alpha}$, 则
$$
\begin{aligned}
\left(\boldsymbol{\beta}, \boldsymbol{A} \boldsymbol{\beta}, \boldsymbol{A}^2 \boldsymbol{\beta}\right) & =\left(\boldsymbol{\alpha}_1+\boldsymbol{\alpha}_2+\boldsymbol{\alpha}_3, \lambda_1 \boldsymbol{\alpha}_1+\lambda_2 \boldsymbol{\alpha}_2+\lambda_3 \boldsymbol{\alpha}_3, \lambda_1^2 \boldsymbol{\alpha}_1+\lambda_2^2 \boldsymbol{\alpha}_2+\lambda_3^2 \boldsymbol{\alpha}_3\right) \\
& =\left(\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3\right)\left(\begin{array}{lll}
1 & \lambda_1 & \lambda_1^2 \\
1 & \lambda_2 & \lambda_2^2 \\
1 & \lambda_3 & \lambda_3^2
\end{array}\right),
\end{aligned}
$$
由于 $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3$ 是不同特征值 $\lambda_1, \lambda_2, \lambda_3$ 对应的特征向量, 故 $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3$ 线性无关, 则 $\left(\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3\right)$ 可逆, 所以 $r\left(\boldsymbol{\beta}, \boldsymbol{A} \boldsymbol{\beta}, \boldsymbol{A}^2 \boldsymbol{\beta}\right)=r\left[\left(\begin{array}{lll}1 & \lambda_1 & \lambda_1^2 \\ 1 & \lambda_2 & \lambda_2^2 \\ 1 & \lambda_3 & \lambda_3^2\end{array}\right)\right]=3$,故 $\boldsymbol{\beta}, \boldsymbol{A} \boldsymbol{\beta}, \boldsymbol{A}^2 \boldsymbol{\beta}$ 线性无关.
（3）【解】
$$
\begin{aligned}
\boldsymbol{A}\left(\boldsymbol{\beta}, \boldsymbol{A} \boldsymbol{\beta}, \boldsymbol{A}^2 \boldsymbol{\beta}\right) & =\left(\boldsymbol{A} \boldsymbol{\beta}, \boldsymbol{A}^2 \boldsymbol{\beta}, \boldsymbol{A}^3 \boldsymbol{\beta}\right)=\left(\boldsymbol{A} \boldsymbol{\beta}, \boldsymbol{A}^2 \boldsymbol{\beta}, 2 \boldsymbol{A} \boldsymbol{\beta}\right) \\
& =\left(\boldsymbol{\beta}, \boldsymbol{A} \boldsymbol{\beta}, \boldsymbol{A}^2 \boldsymbol{\beta}\right)\left(\begin{array}{lll}
0 & 0 & 0 \\
1 & 0 & 2 \\
0 & 1 & 0
\end{array}\right) \\
\boldsymbol{P} & =\left(\boldsymbol{\beta}, \boldsymbol{A} \boldsymbol{\beta}, \boldsymbol{A}^2 \boldsymbol{\beta}\right), \boldsymbol{B}=\left(\begin{array}{lll}
0 & 0 & 0 \\
1 & 0 & 2 \\
0 & 1 & 0
\end{array}\right)
\end{aligned}
$$
则 $\boldsymbol{A P}=\boldsymbol{P B}$, 又由 (2) 知 $\boldsymbol{P}$ 可逆, 故 $\boldsymbol{P}^{-1} \boldsymbol{A P}=\boldsymbol{B}$, 即 $\boldsymbol{A}$ 与 $\boldsymbol{B}$ 相似. 由
$$
|\mu \boldsymbol{E}-\boldsymbol{B}|=\left|\begin{array}{ccc}
\mu & 0 & 0 \\
-1 & \mu & -2 \\
0 & -1 & \mu
\end{array}\right|=\mu\left(\mu^2-2\right)=0,
$$
得 $\boldsymbol{B}$ 的特征值 $\mu_1=0, \mu_2=\sqrt{2}, \mu_3=-\sqrt{2}$, 故 $\boldsymbol{A}$ 的特征值 $\lambda_1=0, \lambda_2=\sqrt{2}, \lambda_3=-\sqrt{2}$.
(4)【证】由 $\left(\boldsymbol{A}^2-2 \boldsymbol{E}\right) \boldsymbol{A} \boldsymbol{\beta}=\boldsymbol{A}^3 \boldsymbol{\beta}-2 \boldsymbol{A} \boldsymbol{\beta}=\mathbf{0}$ 和
$$
\left(\boldsymbol{A}^2-2 \boldsymbol{E}\right) \boldsymbol{A}^2 \boldsymbol{\beta}=\boldsymbol{A}^4 \boldsymbol{\beta}-2 \boldsymbol{A}^2 \boldsymbol{\beta}=\boldsymbol{A}\left(\boldsymbol{A}^3 \boldsymbol{\beta}-2 \boldsymbol{A} \boldsymbol{\beta}\right)=\boldsymbol{0},
$$
知 $\boldsymbol{A} \boldsymbol{\beta}$ 和 $\boldsymbol{A}^2 \boldsymbol{\beta}$ 都是方程组 $\left(\boldsymbol{A}^2-2 \boldsymbol{E}\right) \boldsymbol{x}=\mathbf{0}$ 的解.
由 (2) 知 $\boldsymbol{\beta}, \boldsymbol{A} \boldsymbol{\beta}, \boldsymbol{A}^2 \boldsymbol{\beta}$ 线性无关, 故 $\boldsymbol{A} \boldsymbol{\beta}, \boldsymbol{A}^2 \boldsymbol{\beta}$ 也线性无关.
由 (3) 知, $\boldsymbol{A} \sim \boldsymbol{B}=\left(\begin{array}{lll}0 & 0 & 0 \\ 1 & 0 & 2 \\ 0 & 1 & 0\end{array}\right)$, 则
$$
\boldsymbol{A}^2-2 \boldsymbol{E} \sim \boldsymbol{B}^2-2 \boldsymbol{E}=\left(\begin{array}{ccc}
-2 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{array}\right),
$$
所以
$$
\begin{gathered}
r\left(\boldsymbol{A}^2-2 \boldsymbol{E}\right)=r\left(\boldsymbol{B}^2-2 \boldsymbol{E}\right)=1, \\
3-r\left(\boldsymbol{A}^2-2 \boldsymbol{E}\right)=2 .
\end{gathered}
$$
综上, $\boldsymbol{A} \boldsymbol{\beta}$ 和 $\boldsymbol{A}^2 \boldsymbol{\beta}$ 是方程组 $\left(\boldsymbol{A}^2-2 \boldsymbol{E}\right) \boldsymbol{x}=\mathbf{0}$ 的基础解系.