答案：
(1) $y=\frac{1}{2} x^2-x-4$;
(2) 设 $P D$ 交 $B C$ 于 $H, \because \angle O B C=\angle B C P=45^{\circ}, \therefore P C=P H$ 设 $P\left(t, \frac{1}{2} t^2-t-4\right), \therefore H(t, t-4), D(t, 0)$
$$
\therefore P C+P D=P H+P D=-t^2+3 t+4
$$
$\therefore t=\frac{3}{2}$ 时, $P C+P D$ 取得最大值 $\frac{25}{4}$, 止时 $P\left(\frac{3}{2},-\frac{35}{8}\right)$
(3) 新抛物线解析式为 $y=\frac{1}{2} x^2+4 x+\frac{7}{2}$, $E\left(-\frac{7}{2},-\frac{35}{8}\right), F\left(0, \frac{7}{2}\right)$, 设 $M(-4, m), N\left(n, \frac{1}{2} n^2+4 n+\frac{7}{2}\right)$
① $E F$ 为对角线, $\therefore-4+n=-\frac{7}{2}, \therefore n=\frac{1}{2}, N_1\left(\frac{1}{2}, \frac{45}{8}\right)$;
②$E M$ 为对角线, $n=-\frac{15}{2}, N_2\left(-\frac{15}{2}, \frac{13}{8}\right)$
③$E N$ 为对角线, $n=-\frac{1}{2}, N_3\left(-\frac{1}{2}, \frac{13}{8}\right)$