Omega

since $\omega$ is a cubic root of $1$ therefore $\omega^3 = 1$

In this step, I multiplied certain terms by $1$ :

$X = (\frac{3\omega^2 + (2\omega \times 1) + (7 \times 1)}{2\omega^2 + 7\omega + 3} + \frac{(5 \times 1 )- 8\omega}{8 - 5\omega^2} )^2$

now substitute the ones by $\omega^3$ :

$X = (\frac{3\omega^2 + 2\omega^4 + 7 \omega^3}{2\omega^2 + 7\omega + 3} + \frac{5 \omega^3 - 8\omega}{8 - 5\omega^2} )^2$

Take $\omega^2$ common factor from the first fraction and take $-\omega$ from the second fraction so we have:

$X = (\frac{\omega^2(3 + 2\omega^2 + 7 \omega)}{2\omega^2 + 7\omega + 3} + \frac{-\omega(8-5 \omega^2)}{8 - 5\omega^2})^2$

$=(\omega^2 - \omega)^2$

$=(± \sqrt{3} i)^2 = 3i^2 = \boxed{-3}$

Since $\omega$ and $\omega^2$ are cubic roots of $1$ , $\quad \Rightarrow \omega^2 + \omega + 1 = 0\quad \Rightarrow \omega^2 = - \omega - 1$

Therefore,

$X = \left( \dfrac {3\omega^2 + 2\omega + 7} {2\omega^2 + 7\omega + 3} + \dfrac {5 - 8\omega} {8-5\omega^2} \right)^2$

$\quad = \left( \dfrac {-3-3\omega + 2\omega + 7} {-2-2\omega^2 + 7\omega + 3} + \dfrac {5 - 8\omega} {8+5+5\omega^2} \right)^2$

$\quad = \left( \dfrac {4-\omega} {1+5\omega} + \dfrac {5 - 8\omega} {13+5\omega} \right)^2$

$\quad = \left( \dfrac {(4-\omega) (13+5\omega) + (1+5\omega)(5 - 8\omega)} {(5 - 8\omega)(13+5\omega)} \right)^2$

$\quad = \left( \dfrac {(52+7\omega-5\omega^2) + (5+17\omega-40\omega^2) } {(13+70\omega+25\omega^2) } \right)^2$

$\quad = \left( \dfrac {(57+12\omega) + (45+57\omega) } {-12+45\omega } \right)^2 = \left( \dfrac {(19+4\omega) + (15+19\omega) } {-4+15\omega } \right)^2$

$\quad = \left( \dfrac {34+23\omega} {-4+15\omega } \right)^2 = \dfrac {1156+1564\omega+529\omega^2} {16-120\omega+225\omega^2}$

$\quad = \dfrac {627+1035\omega} {-209-355\omega} = \dfrac {3(209+355\omega)} {-(209+355\omega)} = \boxed{-3}$

Though a long solution can be shortened as below, the solutions by multiplication and division by $\omega$ is by far better.
$\omega ^2+\omega = - 1,~~~\omega ^2 - \omega = - \sqrt3i.\\ 2\omega ^2+1=~= - \sqrt3i , ~~~~2\omega +1= +\sqrt3i$

No need of doing complex maths just try to make numerator like denominator. Yes you only need to remember certain properties likeΠw³=1 andw²=1÷w(can't find omega on keypad