(1) Use mathematical induction.
It is obvious that $\left\{\begin{array}{l}x=a_{1}=1 \\ y=a_{2}=2\end{array}\right.$.
is on the curve. Now assume the integer point $\left(a_{2 k-1}, a_{2 k}\right)$ is on the curve. Then we have
$$a_{2 k-1}^{2}+a_{2 k-1} a_{2 k}-a_{2 k}^{2}+1=0 .$$

By completing the square and transforming, we get
$$\begin{array}{l} \left(a_{2 k-1}+a_{2 k}\right)^{2}+\left(a_{2 k-1}+a_{2 k}\right)\left(a_{2 k-1}+2 a_{2 k}\right) \\ -\left(a_{2 k-1}+2 a_{2 k}\right)^{2}+1=0 . \\ \text { Substituting } a_{n+2}=a_{n+1}+a_{n} \text {, we get } \\ a_{2 k-1}^{2}+a_{2 k+1} a_{2 k+2}-a_{2 k+2}^{2}+1=0 . \end{array}$$

This shows that the integer point $\left(a_{2 k+1}, a_{2 k+2}\right)$ is on the curve $x^{2}+x y-y^{2}+1=0$.
By mathematical induction, the integer points
$$\left(a_{1}, a_{2}\right),\left(a_{3}, a_{4}\right), \cdots,\left(a_{2 k-1}, a_{2 k}\right), \cdots$$

are all on the curve $x^{2}+x y-y^{2}+1=0$.
(2) Decompose $f(x)$ according to the requirements of $g(x)$, so that a common factor $g(x)$ can be factored out, we have
$$\begin{aligned} f(x)= & a_{1} x^{n}+\left(a_{2}-a_{1}\right) x^{n-1}-\left(a_{n-1}+a_{n-2}\right) x \\ & -a_{n-1} \\ = & a_{1} x^{n}+\left(a_{2}-a_{1}\right) x^{n-1} \\ & +\sum_{k=1}^{n-3}\left(a_{k+2}-a_{k+1}-a_{k}\right) x^{n-k-1} \\ & -\left(a_{n-1}+a_{n-2}\right) x-a_{n-1} \\ = & a_{1} x^{n}+a_{2} x^{n-1}+\sum_{k=1}^{n-3} a_{k+2} x^{n-k-1}-a_{1} x^{n-1} \\ & -\sum_{k=1}^{n-3} a_{k+1} x^{n-k-1}-a_{n-1} x-\sum_{k=1}^{n-3} a_{k} x^{n-k-1} \\ & -a_{n-2} x-a_{n-1} \\ = & \sum_{k=1}^{n-1} a_{1} x^{n-k+1}-\sum_{k=1}^{n-1} a_{k} x^{n-k}-\sum_{k=1}^{n-1} a_{k} x^{n-k-1} \\ = & \left(x^{2}-x-1\right) \sum_{k=1}^{n-1} a_{k} x^{n-k-1} . \end{aligned}$$

Since $a_{k}$ are all positive integers, $g(x)=x^{2}-x-1$ divides $f(x)$.