Explain how it solved?
The first one is different from the second.
1) $\displaystyle \frac{(2ab^2 + 6)(-4ab + 2)}{2a^2}$
Take out a common factor of $\displaystyle 2$
$\displaystyle =\frac{2(ab^2 + 3)(-4ab + 2)}{2a^2}$
Cancel the $\displaystyle 2$s in the numerator and denominator
$\displaystyle =\frac{(ab^2 + 3)(-4ab + 2)}{a^2}$
Now you could either expand the numerator, or take out a common factor of $\displaystyle -2$, whichever you think is simpler.
2) $\displaystyle \frac{(2ab^2 + 6) + (-4ab^2 + 2)}{2a^2} = \frac{2ab^2 + 6 - 4ab^2 + 2}{2a^2}$
Collect like terms in the numerator
$\displaystyle =\frac{8 - 2ab^2}{2a^2}$
Take out the common factor of $\displaystyle 2$ in the numerator
$\displaystyle =\frac{2(4 - ab^2)}{2a^2}$
Cancel out the $\displaystyle 2$ in the numerator and denominator
$\displaystyle =\frac{4 - ab2}{a^2}$.