$解：(1)原式 ={\frac {{m}^{2}-9-7} {m+3}}·{\frac {2(m+3)} {m-4}}$
$\ \ \ \ \ \ \ \ \ \ \ ={\frac {(m+4)(m-4)} {m+3}}·{\frac {2(m+3)} {m-4}}$
$\ \ \ \ \ \ \ \ \ \ \ =2m+8$
$當(dāng)m=-1時(shí)，原式=2×(-1)+8=6$
$(2) 原式 ={\frac {3a(a+1)-a(a-1)} {(a+1)(a-1)}}·{\frac {(a+1)(a-1)} {a}}$
$\ \ \ \ \ \ \ ={\frac {2{a}^{2}+4a} {(a+1)(a-1)}}·{\frac {(a+1)(a-1)} {a}}$
$\ \ \ \ \ \ \ ={\frac {a(2a+4)} {(a+1)(a-1)}}·{\frac {(a+1)(a-1)} {a}}$
$\ \ \ \ \ \ \ =2a+4$
$當(dāng)a=8時(shí)，原式=2×8+4=20$
