= Solution
A <survivor function> is $S(t)=\mathbb P(T>t)$. For a nonnegative event time it is nonincreasing and right-continuous, with values between zero and one; a proper finite event time has $S(t)\to0$ as $t\to\infty$. Here $w(0)=1$, $w(t)>0$, $w(t)\to0$, and $w'(t)=-(e^{-t}+6e^{-2t})/4<0$ for $t\geq0$. Extending it by one for $t<0$ supplies a valid continuous survivor function.
The density is $-w'$, so the <hazard function> is
$$
\boxed{h(t)=\frac{e^{-t}+6e^{-2t}}{e^{-t}+3e^{-2t}}
=1+\frac{3e^{-t}}{1+3e^{-t}},\qquad
h'(t)=-\frac{3e^{-t}}{(1+3e^{-t})^2}<0.}
$$
In particular $h(0)=7/4$ and $\boxed{\lim_{t\to\infty}h(t)=1}$. The survivor is a mixture of rate-one and rate-two <exponential distributions> with weights $1/4$ and $3/4$. Among long-term survivors the rate-one component dominates, explaining the limiting hazard. More generally the <hazard derivative for a mixture of exponential distributions> is minus the variance of the component rate among survivors; the decreasing hazard is selection, not an individual component's changing hazard.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-41-mixture-hazard.png]
{title=Survival and declining hazard of a two-rate exponential mixture}
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