答案：
解: (1) 猜想 $y$ 与 $x$ 之问满足一次函数关系，
设一次函数的关系式为 $y=k x+b(k \neq 0)$,
选取 $\left\{\begin{array}{l}x=90, \\ y=600\end{array}\right.$ 和 $\left\{\begin{array}{l}x=100, \\ y=400,\end{array}\right.$ 依题意得 $\left\{\begin{array}{l}90 k+b=600, \\ 100 k+b=400 .\end{array}\right.$ 解得 $\left\{\begin{array}{l}k=-20, \\ b=2400 .\end{array}\right.$
$$
\therefore y=-20 x+2400 \text {. }
$$
经检验, $\left\{\begin{array}{l}x=80, \\ y=800\end{array}\right.$ 和 $\left\{\begin{array}{l}x=110, \\ y=200\end{array}\right.$ 均符合题意.
(2)设获利为 $w$ 元, 依题意, 得 $w=(-20 x+2400)(x-80)$
$$
\begin{aligned}
& =-20 x^2+4000 x-192000=-20(x-100)^2+8000 . \\
& \because a=-20 < 0 \text {, } \\
& \therefore \text { 拋物线开口向下. } \\
& \because \text { 对称轴为直线 } x=100 \text {, } \\
& \text { 又 } \because 80 \leq x \leq 80 \times(1+15 \%) \text {, 即 } 80 \leq x \leq 92, w \text { 随 } x \text { 的增大而增大, } \\
& \therefore \text { 当 } x=92 \text { 时, } w \text { 的最大值为 } 6720 \text { 元. } \\
&
\end{aligned}
$$