It is
not possible.
Consider the remainder of the number when divided by $13$. Initially we start with remainder $3$ ($81\equiv 3 \mod 13$), we have to get remainder $4$ ($82\equiv 4 \mod 13$).
Applying step 1, assuming that $m\equiv 3\mod 13$, we get:
$$
\begin{align}
m &\to m^k \\
3 &\to 3^1, 3^2, 3^3, \ldots \mod 13 \\
3 &\to 3, 9, 1, 3, \ldots \mod 13
\end{align}
$$
Now we know that we can at least get to $1$, $3$, and $9 \mod 13$. Applying the same argument for the two new values, we get:
$$
\begin{align}
1 &\to 1^1, 1^2, 1^3, \ldots \mod 13 \\
1 &\to 1, 1, 1, \ldots \mod 13 \\
9 &\to 9^1, 9^2, 9^3, \ldots \mod 13 \\
9 &\to 9, 3, 1, 9, \ldots \mod 13 \\
\end{align}
$$
(Here we have explicitly shown the argument for $9$, but we could have used the fact that $9=3^2$, so the powers of $9$ must be a subset of the powers of $3$ computed earlier.)
Therefore using step 1 we can get only numbers which are congruent to 1, 3 or 9 modulo 13. Now we verify that this is still true even if we apply step 2.
If we start with the number m and apply step 2, we will get:
$$
\begin{align}
m=100a+b &\to a + 3b
\end{align}
$$
and then:
$$
\begin{align}
100a+b &\to a+3b \mod 13 \\
100a+b &\to 300a + 3b \mod 13 \\
m &\to 3m \mod 13
\end{align}
$$
Thus it is sufficient to consider the action of $m\to 3m$ on the values from earlier ($1$, $3$, and $9$):
$$
\begin{align}
3\times 1 &\to 3 \mod 13 \\
3\times 3 &\to 9 \mod 13 \\
3\times 9 &\to 1 \mod 13 \\
\end{align}
$$
(Again, we could have used the fact that all the values are powers of three instead of calculating all the transformations explicitly.)
Thus the numbers that we get through any number of jumps is always of the form $13k+1$, $13k+3$, or $13k+9$ ($m\equiv1,3,9\mod 13$).
Since $82$ gives remainder $4$ when divided by $13$, it is not reachable.
