解：$(1) $方程组$ \{\begin{array}{l}x+2y=7\\x-y=1\end{array}$的解$ x $与$ y $具有 “邻好关系”
理由如下：
解方程组$ \{\begin{array}{l}x+2y=7\\x-y=1\end{array} $得$ \{\begin{array}{l}x=3\\y=2\end{array}$
∵$|x-y|=1$
∴方程组$ \{\begin{array}{l}x+2y=7\\x-y=1\end{array}$的解$ x $与$ y $具有 “邻好关系”
$(2) $解方程组$ \{\begin{array}{l}2x-y=6\\4x+y=6m\end{array}$得$ \{\begin{array}{l}x=m+1\\y= m-4\end{array} $
∵该方程组的解$ x $与$ y $具有 “邻好关系”
∴$|x-y|=1，$即$ |m+1-(2m-4)|=1，$解得$ m= 4 $或$ 6$
$(3) $解方程组$ \{\begin{array}{l}x+ay=7\\2y-x=5\end{array} $得$ \{\begin{array}{l}x=\frac {14-5a}{a+2}\\y=\frac {12}{a+2}\end{array}$
∵$a $与$x，$$y$都是正整数
∴$\{\begin{array}{l}a=1\\x=3\\y=4\end{array}$或$\{\begin{array}{l}a=2\\x=1\\y=3\end{array}$
∵$|3-4|=1，$$|1-3| \neq 1$
∴当$ a=1 $时，方程组$ \{\begin{array}{l}x+ay=7\\2y-x=5\end{array}$的解$ x $与$ y $具有 “邻好关系”，且方程组的解为$ \{\begin{array}{l}x=3\\y=4\end{array}$
