1. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=1, a_{n}=5 a_{n-1}+3^{n-1}(n \geqslant 2) \text {. }
$$

Find $a_{n}$ and the highest power of 2 that divides $a_{k}\left(k=2^{2019}\right)$
1. Notice that,
$$
\begin{array}{l}
a_{n}=5^{n-1}+5^{n-2} \times 3+\cdots+5 \times 3^{n-2}+3^{n-1} \\
=\frac{1}{2}(5-3)\left(5^{n-1}+5^{n-2} \times 3+\cdots+\right. \\
\left.5 \times 3^{n-2}+3^{n-1}\right) \\
=\frac{1}{2}\left(5^{n}-3^{n}\right) \text {. } \\
\end{array}
$$

When $k=2^{2019}$,
$$
\begin{array}{l}
2 a_{2^{2019}}=5^{2^{2019}}-3^{2^{2019}} \\
=(5-3)(5+3)\left(5^{2}+3^{2}\right) \cdots\left(5^{2^{2018}}+3^{2^{2018}}\right) \\
\Rightarrow a_{2^{2019}} \\
\quad=2^{3}\left(5^{2}+3^{2}\right)\left(5^{2^{2}}+3^{2^{2}}\right) \cdots\left(5^{2^{2018}}+3^{2^{2018}}\right) .
\end{array}
$$

For $m \geqslant 1$,
$$
\begin{array}{l}
5^{2 m} \equiv 1(\bmod 4), 3^{2^{m}} \equiv 1(\bmod 4) \\
\Rightarrow 5^{2^{m}}+3^{2^{m}} \equiv 2(\bmod 4)
\end{array}
$$

Thus, $2^{2018} \times 8\left\|a_{k} \Rightarrow 2^{2021}\right\| a_{k}$.
Therefore, $v_{2}\left(a_{2^{2019}}\right)=2021$.