8. A. If the equation with respect to $x$
$$
x^{2}+k x+\frac{3}{4} k^{2}-3 k+\frac{9}{2}=0
$$

has two real roots $x_{1}$ and $x_{2}$, then $\frac{x_{1}^{2011}}{x_{2}^{2012}}=$
8. A. $-\frac{2}{3}$.

From the problem, we have
$$
\begin{array}{l}
\Delta=k^{2}-4\left(\frac{3}{4} k^{2}-3 k+\frac{9}{2}\right) \geqslant 0 \\
\Rightarrow(k-3)^{2} \leqslant 0 . \\
\text { Also, } (k-3)^{2} \geqslant 0, \text { so } \\
(k-3)^{2}=0 \Rightarrow k=3 .
\end{array}
$$

At this point, the equation is $x^{2}+3 x+\frac{9}{4}=0$.
Solving, we get $x_{1}=x_{2}=-\frac{3}{2}$.
$$
\text { Therefore, } \frac{x_{1}^{2011}}{x_{2}^{2012}}=\frac{1}{x_{2}}=-\frac{2}{3} \text {. }
$$