5. Let $n=1990$, then
$$
\begin{aligned}
& \frac{1}{2^{n}}\left(1-3 C_{n}^{2}+3^{2} C_{n}^{4}-3^{3} C_{n}^{6}+\cdots+3^{994} C_{n}^{1988}-3^{995} C_{n}^{1990}\right) \\
= &
\end{aligned}
$$
5.
$$
\begin{aligned}
\text { Original expression }= & \left(\frac{1}{2}\right)^{1990}-C_{1990}^{2}\left(\frac{1}{2}\right)^{1988}\left(\frac{\sqrt{3}}{2}\right)^{2}+\cdots \\
& +C_{1990}^{1988}\left(\frac{1}{2}\right)^{2}\left(\frac{\sqrt{3}}{2}\right)^{1988}-C_{1990}^{1990}\left(\frac{\sqrt{3}}{2}\right)^{1990}
\end{aligned}
$$

It can be seen that the above expression is the real part of the expansion of
$$
\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{1990}
$$

and
$$
\begin{aligned}
\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{1990} & =\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{1990} \\
& =\left(\cos \frac{1990}{3} \pi+i \sin \frac{1990}{3}\right) \pi
\end{aligned}
$$

Therefore, the answer should be $\cos \frac{1990}{3} \pi$, which is $-\frac{1}{2}$.