7・2 Betya uses a computer in a store, and must pay according to the following standards: for each number input to be multiplied by 3, he has to pay 5 kopecks, and for adding 4 to any number, he has to pay 2 kopecks, but inputting 1 into the computer is free. Betya hopes to calculate from 1 to 1981 with the least amount of money. How much money does he need? And if he wants to calculate to 1982, how much would it cost?
［Solution］(1) We adopt the inverse operation, i.e., perform subtraction by 4 and division by 3 from 1981. When the number is relatively large, it is obviously more cost-effective to use division by 3.
Since 1981 is not divisible by 3, we first subtract 4, and the subsequent operations are as follows:
$$
1981 \xrightarrow{-4} 1977 \xrightarrow{\div 3} 659 \xrightarrow{-4} 655 \xrightarrow{-4} 651 \xrightarrow{\div 3} 217 \xrightarrow{-4} 213 \xrightarrow{\div 3} 71 \xrightarrow{-4} 67 \xrightarrow{-4} 63 \xrightarrow{\div 3} 21 .
$$

There are two ways from 1 to 21:
The first way: $1 \xrightarrow{\times 3} 3 \xrightarrow{+4} 7 \xrightarrow{\times 3} 21$.
This costs $2 \cdot 5 + 2 = 12$ kopecks.
The second way: $1 \xrightarrow{+4} 5 \xrightarrow{+4} 9 \xrightarrow{+4} 13 \xrightarrow{+4} 17 \xrightarrow{+4} 21$.
This costs $2 \cdot 5 = 10$ kopecks.
Clearly, the second way is more cost-effective.
Thus, the operation path from 1 to 1981 is
$$
\begin{aligned}
1 \stackrel{+4}{\rightarrow} 5 \xrightarrow{+4} 9 \xrightarrow{+4} 13 \xrightarrow{+4} 17 \xrightarrow{+4} 21 \xrightarrow{\times 3} 63 \xrightarrow{+4} 67 \xrightarrow{+4} 71 \xrightarrow{\times 3} 213 \xrightarrow{+4} 217 \xrightarrow{\times 3} 651 \xrightarrow{+4} 655 \\
\xrightarrow{+4} 659 \xrightarrow{\times 3} 1977 \xrightarrow{+4} 1981 .
\end{aligned}
$$

It costs a total of $11 \cdot 2 + 4 \cdot 5 = 42$ kopecks.
(2) Since the series of operations on the number is
$$
a_{0}=1, a_{1}, a_{2}, \cdots, a_{k-1}, a_{k}, \cdots, a_{n} .
$$

where
$$
a_{k}=3 a_{k-1}, \quad a_{k}=a_{k-1}+4 .
$$

When $a_{k-1}$ is odd, both $3 a_{k-1}$ and $a_{k-1}+4$ are odd, thus $a_{k}$ is odd.
Since $a_{0}=1$ is odd, all $a_{i} (i=1,2, \cdots)$ are odd, so it is impossible to get 1982.