解：$(1)①$由题意得$\begin{cases}{a-2b-1=-2 } \\{ 4a+4b-1=3} \end{cases}$
解得：$\begin{cases}{ a=\dfrac 13 } \\{ b=\dfrac 23 } \end{cases}$
②∵$a=\frac {1}{3}，$$b=\frac {2}{3}$
∴$T(x，$$y)=\frac {1}{3}x+ \frac {4}{3}y-1$
由题意，得$\begin{cases}{ \dfrac {2}{3}m+\dfrac {4(5-4m)}{3}-1 \leqslant 4①} \\{\dfrac {1}{3}m+\dfrac {4(3-2m)}{3}-1>p② } \end{cases}$
解不等式①得$ m \geqslant \frac {5}{14}；$解不等式②得$ m< \frac {9-3p}{7} $
∵原不等式组恰好有$ 2 $个整数解
∴原不等式组的解集为$ \frac {5}{14} \leqslant m<\frac {9-3p}{7}，$且整数解为$ m=1，$$2 $
∴$2<\frac {9-3p}{7} \leqslant 3，$解得$ -4 \leqslant p<-\frac {5}{3}$
∴$p $的取值范围为$ -4 \leqslant p<-\frac {5}{3}$
$(2)$∵$T(x，$$y)=ax+2by-1，$$T(y，$$x)=ay+2bx-1，$且$T(x，$$y)=T(y，$$x)$
∴$ax+2by-1=ay+2bx-1$
∴$(a-2b)(x-y)=0$
∵对任意实数$x，$$y，$该式都成立
∴$a-2b=0$
∴$a，$$b$之间应满足的关系式为$a=2b$
