Answer 1

Let T_r+1 be the 7th term from the end in the given expression.
Then, we have:
T_r+1 = (9 − 7 + 1) =  3rd term from the beginning
Now, 

\[T_3 = T_{2 + 1} \]
\[ = ^{8}{}{C}_2 (2 x^2 )^{8 - 2} \left( - \frac{3}{2x} \right)^2 \]
\[ = \frac{8 \times 7}{2 \times 1}\left( 64 x^{12} \right)\frac{9}{4 x^2}\]
\[ = 4032 x^{10}\]