2. Let $n$ be a given positive integer, find the smallest positive integer $m$, such that
$$2^{m} \equiv 1\left(\bmod 5^{n}\right)$$
2. This is equivalent to finding $\delta_{5^{n}}(2)$, the answer is: the smallest positive integer $m=4 \times 5^{n-1}$.

We prove the above conclusion by induction on $n$. Direct calculation shows that $\delta_{5}(2)=4$. Now assume $\delta_{5^{n}}(2)=4 \times 5^{n-1}$, then we can set $2^{4 \times 5^{n-1}}=t \cdot 5^{n}+1, t \in \mathbf{N}^{*}$, thus
$$2^{4 \times 5^{n}}=\left(t \cdot 5^{n}+1\right)^{5} \equiv \mathrm{C}_{5}^{1} \cdot t \cdot 5^{n}+1 \equiv 1\left(\bmod 5^{n+1}\right),$$

Therefore, $\delta_{5^{n+1}}(2) \leqslant 4 \times 5^{n}$. Any $m$ that satisfies $2^{m} \equiv 1\left(\bmod 5^{n+1}\right)$ also satisfies $2^{m} \equiv 1(\bmod 5^{n})$, hence $\delta_{5^{n}}(2) \mid \delta_{5^{n+1}}(2)$. Thus, $\delta_{5^{n+1}}(2)=4 \times 5^{n-1}$ or $4 \times 5^{n}$. Since $2^{4 \times 5^{n-1}}=16^{5^{n-1}}=(15+1)^{5^{n-1}}$, using the binomial theorem expansion, we get $5^{n} \|\left(2^{4 \times 5^{n-1}}-1\right)$, so it must be that $\delta_{5^{n+1}}(2)=4 \times 5^{n}$.