解 ：?$(2)$?方程?$ax + by = c $?與它的?$“$?交換系數(shù)方程?$”$?
組成的方程組為?$\begin {cases}ax + by = c\\cx + by = a\end {cases}$?或?$\begin {cases}ax + by = c\\ax + cy = b\end {cases}。$?
?$ $?當(dāng)?$a + b + c = 0$?時(shí)，
對(duì)于?$\begin {cases}ax + by = c\\cx + by = a\end {cases}，$?解得：?$\begin {cases}{x=-1}\\{y=-1}\end {cases}$?
?$ $?對(duì)于?$\begin {cases}ax + by = c\\ax + cy = b\end {cases}，$?解得：?$\begin {cases}{x=-1}\\{y=-1}\end {cases}$?
?$ $?把?$\begin {cases}x=-1\\y =-1\end {cases}$?代入?$mx+ny = p，$?得?$-(m + n)=p。$?
?$ $?所以?$(m + n)m-p(n + p)+2025$?
?$=-pm-pn-p^2+2025$?
?$=-p(m + n)-p^2+2025$?
?$=(-p)^2-p^2+2025 $?
?$= 2025$?
?$ (3)(1 + n)x+2025y = 2m + 2$?的?$“$?交換系數(shù)方
程?$”$?為?$(2m + 2)x+2025y = 1 + n$?或
?$(1 + n)x+(2m + 2)y = 2025。$?
?$ $?因?yàn)?$(10m - t)x+2025y = m + t $?是關(guān)于?$x，y$?的二
元一次方程?$(1 + n)x+2025y = 2m + 2$?的?$“$?交換
系數(shù)方程”，
所以當(dāng)?$(10m - t)x+2025y = m + t $?各系數(shù)與
?$(2m + 2)x+2025y = 1 + n$?各系數(shù)對(duì)應(yīng)相等時(shí)，
得?$\begin {cases}10m - t=2m + 2\\m + t=1 + n\end {cases}，$?解得：?$\begin {cases}{m=\dfrac {t+2}8}\\{n=\dfrac {9t-6}8}\end {cases}$?
∵?$t＜n＜8m，$?
∴?$t＜\frac {9t?6}{8}＜1+2，$?解得?$6＜1＜22(t $?為整數(shù)?$).$?
∴?$8＜1+2＜24，$?
∴若?$m=\frac {t+2}{8}$?為整數(shù)?$，$?必須有?$t+2=16，$?此時(shí)
?$m=2.$?
∴?$t=14.$?
當(dāng)?$t=14$?時(shí)?$，n=\frac {9t?6}{8}=\frac {9×14?6}{8}=\frac {126?6}{8}$?
?$=\frac {120}{8}=15，$?符合題意
∴?$m=2$?
?$ $?當(dāng)?$(10m - t)x+2025y = m + t $?各系數(shù)與
?$(1 + n)x+(2m + 2)y = 2025$?各系數(shù)對(duì)應(yīng)相等時(shí)，
得?$\begin {cases}10m - t=1 + n\\2025=2m + 2\\m + t=2025\end {cases}，$?
解得?$m=\frac {2023}{2}($?不是整數(shù)，舍去）。
綜上，?$m = 2。$?
