Answer
(a) $286~K$
(b) $25.94~years$
Work Step by Step
We are given the function
$$ 283+3\left(1-10^{-0.03 t}\right)$$
(a) We have the limit:
\begin{aligned}
\lim _{t \rightarrow \infty} 283+3\left(1-10^{-0.03 t}\right) &=283+3\left(1-\lim _{t \rightarrow \infty}\left(10^{-0.03 t}\right)\right) \\
&=283+3(1-0) \\
&=286~K
\end{aligned}
(b) We find $t$, when $T=286-0.5$
\begin{aligned}
283+3\left(1-10^{-0.03 t}\right)&\geq 285.5\\
1-10^{-0.03 t} &\geq \frac{2.5}{3}\\
-10^{-0.03 t} &\geq-\frac{1}{6}\\
10^{-0.03 t}& \leq \frac{1}{6}\\
-0.03 t &\leq \log_{10} \frac{1}{6}
\end{aligned}
Then
$$ t \approx 25.94~yrs$$
