$1)$
$(x-2)(x+3)>0$
$→\begin{cases}x-2<0\\x+3>0\\\end{cases}↔\begin{cases}x<2\\x>-3\\\end{cases}$
$→-3<x<2$
$→x∈\{-2;-1;0;1\}$
$2)$
$|a+3|=7$
$→$ \(\left[ \begin{array}{l}a+3=7\\a+3=-7\end{array} \right.\)
$→$ \(\left[ \begin{array}{l}a=4\\a=-10\end{array} \right.\)
Vậy $a∈{4;-10}$
$3)$
C1: $45 - 9 . ( 13 + 5 )$
$=45-9.18$
$=45-162=-117$
C2: $45 - 9 . ( 13 + 5 )$
$=45-9.13-9.5$
$=45-117-45$
$=-117$