6.4 Find the third term of an infinite geometric progression with a common ratio $|q|<1$, the sum of which is $\frac{8}{5}$, and the second term is $-\frac{1}{2}$.
6.4 Let's use formulas (6.10) and (6.6). Then we get the system

$\left\{\begin{array}{l}\frac{b_{1}}{1-q}=\frac{8}{5}, \\ b_{1} q=-\frac{1}{2} .\end{array}\right.$

Further, we have $16 q^{2}-16 q-5=0$, from which $q_{1}=-\frac{1}{4}$, $q_{2}=\frac{5}{4}>1$ (does not fit the condition). Therefore, $b_{3}=\left(-\frac{1}{2}\right)\left(-\frac{1}{4}\right)=\frac{1}{8}$.

Answer: $\frac{1}{8}$.