Answer
a) ${\bf 14.4}/month$
b) ${\bf35}\;\rm month$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The author asked for the cost of operating the 1000-W refrigerator for one month where its compressor works only 20% of the time.
Hence, the time it works during the whole month is given by
$$t=\rm 0.20\left(\dfrac{30\;day}{1\;month}\right)\left(\dfrac{24\;h}{1\;day}\right)=\bf 144\;\rm h/month$$
So it works only for 144 hours in one month.
$${\rm cost}=\rm \left(1\;kW\right)\left(\$0.10\;/kW\;h\right)\left(\dfrac{144\;h}{1\;month}\right)=\$\color{red}{\bf 14.4}/month$$
$$\color{blue}{\bf [b]}$$
Let's assume that the new more-efficient refrigerator (800-W) works only 20% of the time as well.
Hence,
$${\rm cost}=\rm \left(0.8\;kW\right)\left(\$0.10\;/kW\;h\right)\left(\dfrac{144\;h}{1\;month}\right)=\$\color{blue}{\bf 11.52}/month$$
So you will save about $\$2.88$ each month so the extra $\$100$ you pay will take you about $N$ months.
$$N=\dfrac{\$100}{\$2.88}=34.7\approx \color{red}{\bf35}\;\rm month$$
