1. Given the sets $A=\left\{(x, y) \mid y=x^{2}+2 b x+1\right\}, B=\{(x, y) \mid y=2 a(x+b)\}$, where $a, b$ are negative constants, and $A \cap B=\varnothing$, then for any $\theta \in\left[0, \frac{\pi}{2}\right],(a-\cos \theta)^{2}+(b-\sin \theta)^{2}$ the range of values is . $\qquad$
1. $(1,4)$ Detailed explanation: From $A \cap B=\varnothing$, we know that in the equation $x^{2}+2 b x+1=2 a(x+b)$, i.e., $x^{2}+2(b-a) x+1-2 a b=0$, $\Delta<0$ leads to $a^{2}+b^{2}<1$. Therefore, the point $(a, b)$ lies on the part of the circle $a^{2}+b^{2}<1$ (excluding the boundary) in the third quadrant, while the point $(\cos \theta, \sin \theta)$ lies on the quarter circle in the first quadrant of the circle $x^{2}+y^{2}=1$. By drawing a diagram, we can see that the range of $(a-\cos \theta)^{2}+(b-\sin \theta)^{2}$ is $(1,4)$.