We can solve this question, in this way,
- $x^{2}+ 5x+6>0$
$\implies x^{2} + 3x + 2x + 6>0$
$\implies x(x+3) + 2(x+3)>0$
$\implies (x+3)(x+2)>0$
$\implies x>-3 \;\& \;x>-2$
It has infinite range.
- $\left | x+2 \right |>4$
Case$1: x + 2 >4$
$\implies x>2$
Case$2: x + 2 >-4$
$\implies x>-6$
It also has infinite range.
- $9x-7<3x +14$
$\implies 6x<21$
$\implies x < \dfrac{21}{6}$
It also has infinite range.
- $x^{2}- 4x+3<0$
$\implies x^{2}-3x-x+3<0$
$\implies x(x-3)-1(x-3)<0$
$\implies (x-3)(x-1)<0$
Case$1:(x-3)>0\;\&\;(x-1)<0$
$\implies x>3\;\&\;x<1$
It also has infinite range.
Case$1:(x-3)<0\;\&\;(x-1)>0$
$\implies x<3\;\&\;x>1$
$\implies 1<x<3$
It has finite range.
So, the correct answer is $(D).$