Solution. \(H(t) = t^2\), \(S(t) = e^{-t^2}\), \(f(t) = 2te^{-t^2}\). The median solves \(e^{-m^2} = \frac 12\): \(m = \sqrt {\log 2} = 0.833\). (This is a Weibull distribution with \(k = 2\), \(\lambda = 1\).)

Solution. \(S(t + u)/S(t) = e^{-\lambda (t + u)}/e^{-\lambda t} = e^{-\lambda u} = S(u)\). A bulb with a constant hazard that has already burned for \(t\) hours is exactly as likely to last another \(u\) hours as a new one: it does not wear out, and replacing it early gains nothing.

Solution. \(f(t) = 2(1 + t)^{-3}\), so \(h(t) = 2/(1 + t)\), a decreasing hazard. \(\int _t^\infty (1 + u)^{-2}\,du = (1 + t)^{-1}\), so \(m(t) = (1 + t)^{-1}/(1 + t)^{-2} = 1 + t\): the mean residual life is \(1\) when new and \(5\) at age \(4\). Survivors become more reliable, as a decreasing hazard implies.

Solution. \(-\frac {d}{dt}\log S = -S'/S = f/S = h\). For \(S = (1 + t^2)^{-1}\), \(\log S = -\log (1 + t^2)\) and \(h(t) = 2t/(1 + t^2)\), which rises to a maximum of \(1\) at \(t = 1\) and then falls: a hump-shaped hazard, typical of, for example, the risk of relapse after some treatments.

Solution. \(\lambda = 0.2\). \(S(10) = e^{-2} = 0.135\). Median \(\log 2/0.2 = 3.47\) years. The \(90\)th percentile \(t_{0.9}\) has \(S(t_{0.9}) = 0.1\), so \(t_{0.9} = \log 10/0.2 = 11.5\) years.

Solution. The \(30\)-year-old must first survive to \(60\): \[ P(\text {reach } 80 \mid \text {alive at } 30) = P(\text {reach } 60 \mid 30) \times P(\text {reach } 80 \mid 60) = 0.892 \times 0.537 = 0.479 . \] Conditioning on survival to \(60\) removes the risk already passed.