Here are my calculations:
$$P(X<25|X>5)=\frac{P(5<X<25)}{P(X>5)}$$
$$P(5<X<25)=\int_{5}^{25} \lambda e^{-\lambda t} dt=-e^{-\lambda t}\bigg|_{5}^{25}= e^{-\lambda t}\bigg|_{25}^{5}=[1-e^{-25/16}]-[1-e^{-5/16}] \approx 0.52$$
$$P(X<5)=\int_{0}^{5} \lambda e^{-\lambda t} dt=-e^{-\lambda t}\bigg|_{0}^{5}= e^{-\lambda t}\bigg|_{5}^{0}=1-e^{-5/16} \approx 0.27$$
$$P(X>5)=1-P(X<5)=1-0.27=0.73$$
$$P(X<25|X>5)=\frac{0.52}{0.73}\approx 0.71$$
Which has the same probability that $P(X<20)\approx 0.71$. This means, in my opinion, that the peacemaker experiences no wearing out; the probability of a failure before 20 years is the same as the probability of a failure before 25 years. It's counterintuitive though because you would think that any device experiences wearing out during its lifespan. My book says the Weibull distribution is used to model devices that experience wearing out.