$解:(1)代入A(1,t+1),B(t-5,-1)到反比例函數(shù)的圖像上$
$∴t+1=-(t-5)=m,解得t=2,∴A(1,3),B(-3,1),m=3$
$∴反比例函數(shù)的表達式為y_{2}=\frac {3}{x}$
$∵A(1,3),B(-3,-1)在一次函數(shù)y_{1}=kx+b的圖像上$
$∴\begin{cases}{ k+b=3 }\ \\ {\ -3k+b=-1} \end{cases}\begin{cases}{k=1\ }\ \\ { b=2 } \end{cases}$
$∴一次函數(shù)的表達式為y_{1}=x+2$
$(3)∵點(c,p)和點(n,q)在反比例函數(shù)y_{2}=\frac{3}{x}的圖像上$
$∴c= \frac{3}{p},n=\frac{3}{q}$
$∵c=n+1，即\frac{3}{p}=\frac {3}{q}+1$
$∴\frac {q-p}{pq}=\frac{1}{3}$
