8. Arrange $1,2,3, \cdots, 37$ in a row $a_{1}, a_{2}, \cdots, a_{37}$, where $a_{1}=37, a_{2}=1$, and make $a_{1}+a_{2}+\cdots+a_{k}$ divisible by $a_{k+1}$ $(k=1,2, \cdots, 36)$, then $a_{37}=$ $\qquad$ ,$a_{3}=$ $\qquad$
8. 19,2
(1) From the given, we have $a_{37} \mid (a_{1} + a_{2} + \cdots + a_{36})$.
Thus, $a_{37} \mid (a_{1} + a_{2} + \cdots + a_{36} + a_{37})$.
And $a_{1} + a_{2} + \cdots + a_{36} + a_{37} = 1 + 2 + 3 + \cdots + 37 = \frac{37 \times (1 + 37)}{2} = 37 \times 19$.
Therefore,
$$
a_{37} \mid (37 \times 19).
$$

Since $a_{37} \neq 1, 37$, it follows that $a_{37} = 19$.
(2) From the given, $a_{3} \mid (a_{1} + a_{2})$, i.e., $a_{3} \mid 38$, and $a_{3} \neq 1, 19, 38$, so $a_{3} = 2$.