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Given the probability density after $3$ years is proportional to $\frac{t^2}{(1+t^2)^2}$ then we must have that:

\begin{align*}
&& 1 &= A \int_0^{\infty} \frac{t^2}{(1+t^2)^2} \, \d t \\
&&&= A \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{\infty} + \frac{A}2 \int_0^{\infty} \frac{1}{1+t^2} \d t \\
&&&= \frac{A}{2} \frac{\pi}{2} \\
\Rightarrow && A &= \frac{4}{\pi}
\end{align*}

In order to fail a test on the fifth anniversary, there are two possibilities for when we went faulty. We could have gone faulty before $4$ years, got lucky once and then failed the second test, or gone faulty in the next year and then failed the first test.

\begin{align*}
\P(\text{rusty before } 4 \text{ years}) &=\frac{4}{\pi} \int_0^1 \frac{t^2}{(1+t^2)^2} \d t \\
&=  \frac{4}{\pi} \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{1} +  \frac{2}{\pi} \int_0^{1} \frac{1}{1+t^2} \d t \\
&= -\frac{1}{\pi} + \frac{2}{\pi} \frac{\pi}{4} \\
&= \frac12 - \frac{1}{\pi} \\
&\approx 0.181690\cdots \\
\\
\P(\text{rusty before } 5 \text{ years}) &=\frac{4}{\pi} \int_0^1 \frac{t^2}{(1+t^2)^2} \d t \\
&=  \frac{4}{\pi} \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{2} +  \frac{2}{\pi} \int_0^{2} \frac{1}{1+t^2} \d t \\
&= -\frac{4}{5\pi} + \frac{2}{\pi} \tan^{-1} 2 \\
&\approx 0.450184\cdots \\
\end{align*}

Therefore:

\begin{align*}
\P(\text{fails 5th anniversary}) &= \P(\text{rusty before } 4 \text{ years})  \P(\text{pass one, fail other}) + \\
& \quad \quad +   \P(\text{rusty between 4 and 5 years}) \P(\text{fail}) \\
&= 0.181690\cdots \cdot \frac{1}{4} + \frac{1}{2} ( 0.450184\cdots- 0.181690\cdots) \\
&= \frac{1}{2} 0.450184\cdots - \frac{1}{4} 0.181690\cdots \\
&= 0.1796688\cdots \\
&= 18.0\%\,\, (3\text{ s.f.})
\end{align*}

We also must have that:

\begin{align*}
\P(\text{rusty at 4 years}|\text{destroyed at 5}) &= \frac{\P(\text{rusty at 4 years and destroyed at 5})}{\P(\text{destroyed at 5})} \\
&= \frac{0.181690\cdots \cdot \frac{1}{4}}{\frac{1}{2} 0.450184\cdots - \frac{1}{4} 0.181690\cdots} \\
&= 0.252811\cdots \\
&= 25.3\%\,\,(3\text{ s.f.})
\end{align*}

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