Let's calculate the following powers to an accuracy of four decimal places:
a) $1.02^{30}$
b) $0.996^{13}$
$$
\begin{aligned}
& \text { a) } 1.02^{30}=\left(1+2 \cdot 10^{-2}\right)^{30}=1+\binom{30}{1} 2 \cdot 10^{-2}+1+\binom{30}{2} 2^{2} \cdot 10^{-4}+ \\
& +\binom{30}{3} 2^{3} \cdot 10^{-6}+\binom{30}{4} 2^{4} \cdot 10^{-8}+\binom{30}{5} 2^{5} \cdot 10^{-10}+\binom{30}{6} 2^{6} \cdot 10^{-12}+\ldots
\end{aligned}
$$

Since the ratio of the $(k+1)$-th term to the $k$-th term is

$$
\binom{30}{k} 2^{k} \cdot 10^{-2 k}:\binom{30}{k-1} 2^{k-1} \cdot 10^{-2 k+2}=\frac{31-k}{k} \cdot \frac{2}{10^{2}}=\frac{31-k}{50 k}<\frac{1}{12}, \text { if } k \geqq 7
$$

the terms decrease constantly in such a way that if we neglect the last 24 out of 31 terms, the error (crudely estimated) is less than twice the 8th term (since $24 \cdot \frac{1}{12}=$ 2):

$$
\binom{30}{7} 2^{7} \cdot 10^{-14} \cdot 2=2035800 \cdot 256 \cdot 10^{-14}<3 \cdot 10^{6} \cdot 3 \cdot 10^{2} \cdot 10^{-14}=0.000009
$$

Thus, the error does not affect the 4th decimal place. Therefore, it is sufficient to calculate the first 7 terms, and even each term to 5 decimal places:

$$
\begin{aligned}
& 1 \\
& 0.6 \\
& 0.174 \\
& 0.03248 \\
& 0.00438 \\
& 0.00046 \\
& \frac{0.00004}{1.81136 \approx 1.8114} \\
& \text { b) } 0.996^{13}=\left(1-4 \cdot 10^{-3}\right)^{13}=1-\binom{13}{1} 4 \cdot 10^{-3}+\binom{13}{2} 4^{2} \cdot 10^{-6}-\ldots
\end{aligned}
$$

Since the terms are constantly decreasing and alternating in sign, the error is less than the absolute value of the first neglected term. The fourth term

$$
\binom{13}{3} 4^{3} \cdot 10^{-9}=286 \cdot 64 \cdot 10^{-9}=0.000018304
$$

does not affect the 4th decimal place, and thus

$$
1-0.052+0.001248=1.001248-0.052=0.949248 \approx 0.9492
$$