jmc

Rotating $360^{\circ }$ is the same as doing nothing, so rotating $3645^{\circ }$ is the same as rotating $3645^{\circ }-10\cdot 360^{\circ }=45^{\circ }$. Therefore, $\tan (-3645^{\circ })=\tan (-45^{\circ })$.

Rotating $45^{\circ }$ clockwise is the same as rotating $360^{\circ }-45^{\circ }=315^{\circ }$ counterclockwise, so $\tan (-45^{\circ })=\tan (360^{\circ }-45^{\circ })=\tan 315^{\circ }$.

Let $P$ be the point on the unit circle that is $315^{\circ }$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.



Triangle $POD$ is a 45-45-90 triangle, so $DO=OP=\frac{\sqrt{2}}{2}$. Therefore, the coordinates of $P$ are $\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$, so $\tan (-3645^{\circ })=\tan (-45^{\circ })=\tan 315^{\circ }=\frac{\sin 315^{\circ }}{\cos 315^{\circ }}=\frac{-\sqrt{2}/2}{\sqrt{2}/2}=\boxed{-1}$.