2. Given $\odot C:(x-1)^{2}+(y-2)^{2}=25$ and the line $l:(2 a+1) x+(a+1) y=4 a+4 b(a, b \in \mathbf{R})$. If for any real number $a$, the line $l$ always intersects with $\odot C$, then the range of values for $b$ is $\qquad$.
2. $\left[\frac{3-\sqrt{5}}{4}, \frac{3+\sqrt{5}}{4}\right]$.

The center of the circle $C(1,2)$, radius $r=5$.
By the distance formula from a point to a line, the distance from the center $C$ to the line $l$ is
$$
\begin{aligned}
d & =\frac{|(2 a+1)+2(a+1)-4 a-4 b|}{\sqrt{(2 a+1)^{2}+(a+1)^{2}}} \\
& =\frac{|4 b-3|}{\sqrt{5 a^{2}+6 a+2}} .
\end{aligned}
$$

Since the line $l$ always has a common point with the circle $C$, for any $a \in \mathbf{R}$, we have $d \leqslant 5$ always holds, i.e.,
$$
|4 b-3| \leqslant 5 \sqrt{5 a^{2}+6 a+2}
$$

holds for any $a \in \mathbf{R}$. It is easy to get
$$
\left(5 a^{2}+6 a+2\right)_{\min }=\frac{1}{5} .
$$

Therefore, $|4 b-3| \leqslant \sqrt{5}$.
Solving this inequality gives $b \in\left[\frac{3-\sqrt{5}}{4}, \frac{3+\sqrt{5}}{4}\right]$.