Denote $f(x) = \frac{a x^2 + b}{\sqrt{x^2+1}}$. It is easy to see that $a > 0$. By $f(0) = b$, we see that $b \ge 3$.

a. If $b - 2a \ge 0$,
$$f(x) = \frac{a x^2 + b}{\sqrt{x^2 + 1}} = a \sqrt{x^2 + 1} + \frac{b-a}{\sqrt{x^2 + 1}} \ge 2\sqrt{a(b-a)} = 3,$$
equality holds if $a \sqrt{x^2+1} = \frac{b-a}{\sqrt{x^2+1}}$, that is, if $x = \pm\sqrt{\frac{b-2a}{a}}$.
The value of $a$ for given $b$ is $a = \frac{b - \sqrt{b^2 - 9}}{2}$, especially when $b = 3$, $a = \frac{3}{2}$.

b. If $b - 2a < 0$, let $\sqrt{x^2 + 1} = t$ ($t \ge 1$). $f(x) = g(t) = a t + \frac{b-a}{t}$ is monotonically increasing when $t \ge 1$, so,
$$\min_{x \in \mathbb{R}} f(x) = g(1) = a + b - a = b = 3, \text{ when } a > \frac{3}{2}.$$

Summing up, we get (1) the range of $b$ is $[3, +\infty)$.
(2) If $b = 3$, then $a \ge \frac{3}{2}$; if $b > 3$, then $a = \frac{b - \sqrt{b^2 - 9}}{2}$.

Let $f(x) = \frac{a x^2 + b}{\sqrt{x^2+1}}$. It is easy to see that $a > 0$. Since $\min_{x \in \mathbb{R}} \frac{a x^2 + b}{\sqrt{x^2+1}} = 3$, and $f(0) = b$, we see that $b \ge 3$.
$$f'(x) = \frac{a x\left(x^2 - \frac{b-2a}{a}\right)}{\left(x^2 + 1\right)^{3/2}}.$$

a. If $b - 2a \le 0$, let $f'(x) = 0$, we have the solution $x_0 = 0$, and if $x < 0$, then $f'(x) < 0$; and if $x > 0$, then $f'(x) > 0$.
$f(0) = b$ is the minimal value. Hence $b = 3$, and $a \ge \frac{b}{2}$.

b. If $b - 2a > 0$, let $f'(x) = 0$, we have the solutions $x_0 = 0$, $x_{1,2} = \pm\sqrt{\frac{b-2a}{a}}$.
It is easy to see $f(0) = b$ is not the minimal value, which implies $b > 3$; and $f(x_{1,2})$ is the minimal value
$$\begin{align*} f(x_{1,2}) &= \frac{a \cdot \frac{b-2a}{a} + b}{\sqrt{\frac{b-2a}{a} + 1}} = 3 \Rightarrow 2\sqrt{a(b-a)} = 3 \\ \Rightarrow a^2 - ab + \frac{9}{4} &= 0 \Rightarrow a = \frac{b - \sqrt{b^2 - 9}}{2}, \end{align*}$$
that is, $b > 3$ and $a = \frac{b - \sqrt{b^2 - 9}}{2}$.

Summing up, we get
(1) the range of $b$ is $[3, +\infty)$.
(2) If $b = 3$, then $a \ge \frac{3}{2}$; if $b > 3$, then $a = \frac{b - \sqrt{b^2 - 9}}{2}$.