5. Rearrange the 37 different natural numbers $1,2,3, \cdots, 37$ into a sequence, denoted as $a_{1}, a_{2}, \cdots, a_{37}$, where $a_{1}=37, a_{2}=1$, and ensure that $a_{1}+a_{2}+\cdots+a_{k}$ is divisible by $a_{k+1}$ $(k=1,2, \cdots, 36)$. Find $a_{3}=? a_{37}=$?
【Analysis】Obviously, the sum of these 37 numbers is a multiple of $a_{37}$, so the total sum $37 \times 19$ is a multiple of $a_{37}$, from which we can find the value of $a_{37}$. For $a_{3}$, 38 is a multiple of $a_{3}$, from which we can find the value of $a_{3}$, solving the problem.

【Solution】Solution: The sum of these 37 numbers is a multiple of $a_{37}$, so the total sum $37 \times 19$ is a multiple of $a_{37}$, so $a_{37}$ $=19$;

For $a_{3}, a_{3}$ can divide $a_{1}+a_{2}=37+1=38$, so 38 is a multiple of $a_{3}$, so $a_{3}=2$.