(1) 解：
当$n=1$时，
$\begin{aligned}&\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^1-(\frac{1-\sqrt{5}}{2})^1]\\=&\frac{1}{\sqrt{5}}×\frac{1+\sqrt{5}-1+\sqrt{5}}{2}\\=&\frac{1}{\sqrt{5}}×\sqrt{5}\\=&1\end{aligned}$
当$n=2$时，
$\begin{aligned}&\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^2-(\frac{1-\sqrt{5}}{2})^2]\\=&\frac{1}{\sqrt{5}}×(\frac{1+\sqrt{5}}{2}+\frac{1-\sqrt{5}}{2})×(\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2})\\=&\frac{1}{\sqrt{5}}×1×\sqrt{5}\\=&1\end{aligned}$
所以第1个数和第2个数均为1。
(2) 证明：
$\begin{aligned}x_{n+1}-x_n&=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1}]-\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n]\\&=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^{n+1}+(\frac{1-\sqrt{5}}{2})^n]\\&=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^n(\frac{1+\sqrt{5}}{2}-1)-(\frac{1-\sqrt{5}}{2})^n(\frac{1-\sqrt{5}}{2}-1)]\\&=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^n×\frac{\sqrt{5}-1}{2}-(\frac{1-\sqrt{5}}{2})^n×\frac{-\sqrt{5}-1}{2}]\\&=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^n×\frac{\sqrt{5}-1}{2}+(\frac{1-\sqrt{5}}{2})^n×\frac{\sqrt{5}+1}{2}]\\&=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n-1}×\frac{1+\sqrt{5}}{2}×\frac{\sqrt{5}-1}{2}+(\frac{1-\sqrt{5}}{2})^{n-1}×\frac{1-\sqrt{5}}{2}×\frac{\sqrt{5}+1}{2}]\\&=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n-1}×\frac{5-1}{4}+(\frac{1-\sqrt{5}}{2})^{n-1}×\frac{5-1}{4}]\\&=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n-1}-(\frac{1-\sqrt{5}}{2})^{n-1}]\\&=x_{n-1}\end{aligned}$
即$x_{n+1}-x_n=x_{n-1}$（$n$为正整数，且$n≥2$）。
