Answer: It will increase by $\frac{11}{7}$ times.

Solution. Let the common ratio of the geometric progression be $q$. Since all its terms are positive, $q>0$. If $q=1$, then $S=3000 b_{1}$, and when the terms with numbers divisible by 3 are increased by 40 times, we get the sum $S+39\left(b_{3}+b_{6}+\ldots+b_{3000}\right)=S+39 \cdot \frac{S}{3}=14 S$. This contradicts the condition, so $q \neq 1$, which means the sum of the first $n$ terms of the progression can be calculated using the formula $S_{n}=b_{1} \cdot \frac{1-q^{n}}{1-q}$. In particular, $S=b_{1} \frac{1-q^{3000}}{1-q}$.

When the terms $b_{3}, b_{6}, \ldots b_{3000}$ are increased by 40 times, we get the sum

$S+39\left(b_{3}+b_{6}+\ldots+b_{3000}\right)=S+39 b_{3} \frac{1-\left(q^{3}\right)^{1000}}{1-q^{3}}=S+39 b_{1} q^{2} \frac{1-q^{3000}}{(1-q)\left(1+q+q^{2}\right)}=S+\frac{39 q^{2} S}{1+q+q^{2}}$, which equals $5 S$. Therefore, $\frac{39 q^{2}}{1+q+q^{2}}=4,35 q^{2}-4 q-4=0, q=\frac{2}{5}$ or $q=-\frac{2}{7}$. The positive value of $q$ is suitable, i.e., $q=\frac{2}{5}$.

If all terms of the progression with even numbers are tripled, we get

$S+2\left(b_{2}+b_{4}+\ldots+b_{3000}\right)=S+2 \cdot b_{2} \cdot \frac{1-\left(q^{2}\right)^{1500}}{1-q^{2}}==S+2 b_{1} q \frac{1-q^{3000}}{(1-q)(1+q)}=S+\frac{2 q}{1+q} S=\frac{11}{7} S$.

Thus, the sum $S$ will increase by $\frac{11}{7}$ times.
2. [4 points] Given a geometric progression $b_{1}, b_{2}, \ldots, b_{3000}$, all of whose terms are positive, and their sum is $S$. It is known that if all its terms with indices divisible by 3 (i.e., $b_{3}, b_{6}, \ldots, b_{3000}$) are increased by 40 times, the sum $S$ will increase by 5 times. How will $S$ change if all its terms in even positions (i.e., $b_{2}, b_{4}, \ldots, b_{3000}$) are increased by 3 times?