3. Given $F(a, \theta)=\frac{a^{2}+2 a \sin \theta+2}{a^{2}+2 a \cos \theta+2}, a, \theta \in \mathbf{R}$. $\alpha \neq 0$. Then, for any $a, 2(a, \theta)$, the maximum and minimum values are respectively

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3. Given $F(a, \theta)=\frac{a^{2}+2 a \sin \theta+2}{a^{2}+2 a \cos \theta+2}, a, \theta \in \mathbf{R}$. $\alpha \neq 0$. Then, for any $a, 2(a, \theta)$, the maximum and minimum values are respectively
$3.2+\sqrt{3}, 2-\sqrt{3}$.
Let $t=F(a, \theta)$, then
$2 ta \cos \theta - 2 a \sin \theta + (t-1)(a^2 + 2) = 0$.
Therefore, the line $2 a t x - 2 a y + (t-1)(a^2 + 2) = 0$ intersects
the circle $x^2 + y^2 = 1$ at a common point $(\cos \theta, \sin \theta)$. Thus, we have
$\frac{|t-1|(a^2 + 2)}{2|a| \sqrt{t^2 + 1}} \leq 1$. (Equality holds when $\tan \theta = -\frac{1}{t}$)
This implies $\frac{|t-1|}{\sqrt{t^2 + 1}} \leq \frac{2|a|}{a^2 + 2} \leq \frac{2|a|}{2 \sqrt{2}|a|} = \frac{1}{\sqrt{2}}$.
Thus, we have $t^2 - 4t + 1 \leq 0$.
Solving this, we get $2 - \sqrt{3} \leq t \leq 2 + \sqrt{3}$.
Equality holds when $a = \pm \sqrt{2}, \theta = \arctan \frac{1}{2 \pm \sqrt{3}}$.
Therefore, the maximum value of $F(a, \theta)$ is $2 + \sqrt{3}$, and the minimum value is $2 - \sqrt{3}$.