5. Let $n=1990$, then $\frac{1}{2^{n}}\left(1-3 \mathrm{C}_{n}^{2}+3^{2} \mathrm{C}_{n}^{4}-3^{3} \mathrm{C}_{n}^{6}+\cdots+3^{994} \mathrm{C}_{n}^{1988}-3^{995} \mathrm{C}_{n}^{1990}\right)=$
$-\frac{1}{2}$
5.【Analysis and Solution】The original expression $=\left(\frac{1}{2}\right)^{1990}-\mathrm{C}_{1990}^{2}\left(\frac{1}{2}\right)^{1990}\left(\frac{\sqrt{3}}{2}\right)^{2}+\cdots+\mathrm{C}_{1990}^{1998}\left(\frac{1}{2}\right)^{2}\left(\frac{\sqrt{3}}{2}\right)^{1988}-\mathrm{C}_{1990}^{1990}$ $\left(\frac{\sqrt{3}}{2}\right)^{1990}$, it can be seen that the above expression is $\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{1990}=\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{1990}$ $=\left(\cos \frac{1990}{3} \pi+i \sin \frac{1990}{3} \pi\right)$.
Therefore, the answer should be $\cos \frac{1990}{3} \pi$, which is $-\frac{1}{2}$.
【Extension】Solving this problem requires observational skills.
$\frac{1}{2^{n}}\left(1-3 \mathrm{C}_{n}^{2}+3^{2} \mathrm{C}_{n}^{4}-3^{3} \mathrm{C}_{n}^{6}+\cdots 3^{994} \mathrm{C}_{n}^{1988}-3^{999} \mathrm{C}_{n}^{1990}\right)$, after careful analysis, we find that it is actually the imaginary part of $\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{1990}$, thus breaking through the problem. In the 2001 competition, there was a question with the same solving approach:
“If $\left(1+x+x^{2}\right)^{1000}$ is expanded as $a_{0}+a_{1} x+a_{2} x_{2}+\cdots+a_{2000} x^{2000}$, then the value of $a_{0}+a_{3}+a_{6}+a_{9}+\cdots a_{1998}$ is
A. $3^{333}$
B. $3^{666}$
C. $3^{999}$
D. $3^{2001}$”