Answer
Cold-reservoir temperature= $-33^{\circ}C$
Hot-reservoir temperature= $47^{\circ}C$
Work Step by Step
For a carnot engine, efficiency is
$\eta_{carnot}=1-\frac{T_{C}}{T_{H}}$
where $T_{C}$ is the temperature in Kelvin of cold reservoir and $T_{H}$ is the temperature of hot reservoir.
Given: $\eta=0.25$, $T_{H}=T_{C}+80$.
$\implies 0.25=1-\frac{T_{C}}{T_{C}+80}$
$\implies 0.25=\frac{80}{T_{C}+80}$
$\implies 0.25T_{C}+20=80$
$\implies 0.25T_{C}=80-20=60$
Or $T_{C}=\frac{60}{0.25}=240\,K=(240-273)^{\circ}C=-33^{\circ}C$
$T_{H}=T_{C}+80\,K=(240+80)\,K=320\,K$
$=(320-273)^{\circ}C=47^{\circ}C$