1. Given $\min _{x \in R} \frac{a x^{2}+b}{\sqrt{x^{2}+1}}=3$.
(1) Find the range of $b$;
(2) For a given $b$, find $a$.
1. Solution 1 (1) Let $f(x)=\frac{a x^{2}+b}{\sqrt{x^{2}+1}}$.
Then $f(0)=b \geqslant 3$. Furthermore, $a>0$.
(i) When $b-2 a \geqslant 0$,
$$
\begin{array}{l}
f(x)=\frac{a x^{2}+b}{\sqrt{x^{2}+1}} \\
=a \sqrt{x^{2}+1}+\frac{b-a}{\sqrt{x^{2}+1}} \\
\geqslant 2 \sqrt{a(b-a)}=3 .
\end{array}
$$

The equality holds if and only if $a \sqrt{x^{2}+1}=\frac{b-a}{\sqrt{x^{2}+1}}$, i.e., $x=$ $\pm \sqrt{\frac{b-2 a}{a}}$.
At this point, $a=\frac{b-\sqrt{b^{2}-9}}{2}$.
In particular, when $b=3$, $a=\frac{3}{2}$.
(ii) When $b-2 a<0$,
$$
a>\frac{3}{2}.
$$
In summary, the range of $b$ is $[3,+\infty)$,
(2) When $b=3$, $a \geqslant \frac{3}{2}$;

When $b>3$, $a=\frac{b-\sqrt{b^{2}-9}}{2}$.
Solution 2 (1) Let $f(x)=\frac{a x^{2}+b}{\sqrt{x^{2}+1}}$.
Since $\min _{x \in \mathbb{R}} \frac{a x^{2}+b}{\sqrt{x^{2}+1}}=3$, and $f(0)=b$, thus, $b \geqslant 3$.
It is easy to see that $a>0, f^{\prime}(x)=\frac{a x\left(x^{2}-\frac{b-2 a}{a}\right)}{\left(x^{2}+1\right)^{\frac{3}{2}}}$.
(i) When $b-2 a \leqslant 0$, let $f^{\prime}(x)=0$.

Solving gives $x_{0}=0$.
When $x<0$, $f^{\prime}(x)<0$; when $x>0$, $f^{\prime}(x)>0$.
Thus, $f(0)=b$ is the minimum value.
Therefore, $b=3$.
(ii) When $b-2 a>0$, let $f^{\prime}(x)=0$.

Solving gives $x_{0}=0, x_{1,2}= \pm \sqrt{\frac{b-2 a}{a}}$.
At this point, it is easy to see that
$f(0)=b$ is not the minimum value
$\Rightarrow b>3$, and $f\left(x_{1,2}\right)$ is the minimum value
$$
\begin{array}{l}
\Rightarrow f\left(x_{1,2}\right)=\frac{a \cdot \frac{b-2 a}{a}+b}{\sqrt{\frac{b-2 a}{a}+1}}=3 \\
\Rightarrow 2 \sqrt{a(b-a)}=3 \\
\Rightarrow a^{2}-a b+\frac{9}{4}=0 \\
\Rightarrow a=\frac{b-\sqrt{b^{2}-9}}{2} .
\end{array}
$$

In summary, the range of $b$ is $[3,+\infty)$.
(2) When $b=3$, $a \geqslant \frac{3}{2}$;

When $b>3$, $a=\frac{b-\sqrt{b^{2}-9}}{2}$.