3. The heat engine "Lomonosov" 12341 receives heat on the segment $1-2$ and gives it off on the segment $3-4$. Therefore,

$$
\eta_{0}=1-\frac{Q_{34}}{Q_{12}}
$$

(here and below, all heats are assumed to be positive). The engine "Avogadro" 1231 also receives heat on 12, but gives it off on 31, so

$$
\eta_{1}=1-\frac{Q_{13}}{Q_{12}}
$$

The engine "Boltzmann" 1341 receives heat on 13 and gives it off on 34, hence

$$
\eta_{2}=1-\frac{Q_{34}}{Q_{13}}
$$

Combining the three equations above, we get

$$
\eta_{2}=\frac{\eta_{0}-\eta_{1}}{1-\eta_{1}}
$$

$\eta_{1}<\eta_{0}$, because the engine 1231 performs less work with the same amount of heat received, and $\eta_{2}<\eta_{0}$ for all values of $\eta_{0}, \eta_{1}<1$, i.e., for all permissible values. Given that $\eta_{1}=(1-0.01 \alpha) \eta_{0}$, we obtain the answer:
Answer: $\eta_{2}=\frac{\alpha}{100-(100-\alpha) \eta_{0}} ;$ for all.