1. Given $a \in \mathbf{Z}$, and $x^{6}-33 x+20$ can be divided by $x^{2}-x+a$, then the value of $a$ is
1. Let $x^{6}-33 x+20=\left(x^{2}-x+a\right) \cdot g(x)$, where $g(x)=x^{4}+b x^{3}+c x^{2}+d x+e$. By comparing the constant terms on both sides, we know $a e=20$. Therefore, $a= \pm 1, \pm 2, \pm 4, \pm 5$.

By comparing the coefficients of the linear term, we know $a d-e=-33$. If $a= \pm 2$, then by $a e=20$, we know $e$ is even, but by $a d-e=-33$, we know $e$ is odd. This is a contradiction. Similarly, $a= \pm 5$ and $a=-4$ are also not possible. Therefore, $a= \pm 1$ or 4. By checking each case, we find that $a= \pm 1$ does not satisfy the conditions, so $a=4$.