If $\min \left \{ \frac{ax^2+b}{\sqrt{x^2+1}} \mid x \in \mathbb{R}\right \} = 3$, then (1) Find the range of $b$; (2) for every given $b$, find $a$.
1. **Finding the range of \( b \):**

   Given the function \( f(x) = \frac{ax^2 + b}{\sqrt{x^2 + 1}} \), we need to find the minimum value of this function over all \( x \in \mathbb{R} \). We are given that this minimum value is 3.

   First, evaluate the function at \( x = 0 \):
   \[
   f(0) = \frac{a \cdot 0^2 + b}{\sqrt{0^2 + 1}} = b
   \]
   Since the minimum value of the function is 3, we have:
   \[
   b \geq 3
   \]

2. **Finding \( a \) for every given \( b \):**

   To find \( a \), we need to ensure that the minimum value of the function \( f(x) \) is 3. This involves finding the critical points of the function and ensuring that the minimum value at these points is 3.

   The function \( f(x) = \frac{ax^2 + b}{\sqrt{x^2 + 1}} \) can be analyzed by finding its derivative and setting it to zero to find the critical points.

   The derivative of \( f(x) \) is:
   \[
   f'(x) = \frac{d}{dx} \left( \frac{ax^2 + b}{\sqrt{x^2 + 1}} \right)
   \]
   Using the quotient rule:
   \[
   f'(x) = \frac{(2ax \sqrt{x^2 + 1}) - (ax^2 + b) \cdot \frac{x}{\sqrt{x^2 + 1}}}{(x^2 + 1)}
   \]
   Simplifying the numerator:
   \[
   f'(x) = \frac{2ax(x^2 + 1) - (ax^2 + b)x}{(x^2 + 1)^{3/2}}
   \]
   \[
   f'(x) = \frac{2ax^3 + 2ax - ax^3 - bx}{(x^2 + 1)^{3/2}}
   \]
   \[
   f'(x) = \frac{ax^3 + (2a - b)x}{(x^2 + 1)^{3/2}}
   \]
   Setting \( f'(x) = 0 \):
   \[
   ax^3 + (2a - b)x = 0
   \]
   \[
   x(ax^2 + 2a - b) = 0
   \]
   This gives us two cases:
   \[
   x = 0 \quad \text{or} \quad ax^2 + 2a - b = 0
   \]

   For \( x = 0 \):
   \[
   f(0) = b
   \]
   Since \( f(0) = 3 \), we have \( b = 3 \).

   For \( ax^2 + 2a - b = 0 \):
   \[
   ax^2 = b - 2a
   \]
   \[
   x^2 = \frac{b - 2a}{a}
   \]
   Substituting \( x^2 = \frac{b - 2a}{a} \) into \( f(x) \):
   \[
   f(x) = \frac{a \left( \frac{b - 2a}{a} \right) + b}{\sqrt{\left( \frac{b - 2a}{a} \right) + 1}}
   \]
   Simplifying:
   \[
   f(x) = \frac{b - 2a + b}{\sqrt{\frac{b - 2a + a}{a}}}
   \]
   \[
   f(x) = \frac{2b - 2a}{\sqrt{\frac{b - a}{a}}}
   \]
   \[
   f(x) = \frac{2b - 2a}{\sqrt{\frac{b - a}{a}}}
   \]
   For the minimum value to be 3:
   \[
   3 = \frac{2b - 2a}{\sqrt{\frac{b - a}{a}}}
   \]
   Solving for \( a \):
   \[
   3 \sqrt{\frac{b - a}{a}} = 2b - 2a
   \]
   \[
   3 \sqrt{\frac{b - a}{a}} = 2(b - a)
   \]
   \[
   3 \sqrt{\frac{b - a}{a}} = 2(b - a)
   \]
   Squaring both sides:
   \[
   9 \frac{b - a}{a} = 4(b - a)^2
   \]
   \[
   9(b - a) = 4a(b - a)^2
   \]
   \[
   9(b - a) = 4a(b^2 - 2ab + a^2)
   \]
   \[
   9(b - a) = 4ab^2 - 8a^2b + 4a^3
   \]
   \[
   9(b - a) = 4ab^2 - 8a^2b + 4a^3
   \]
   Solving this equation for \( a \) in terms of \( b \):
   \[
   a = \frac{b - \sqrt{b^2 - 9}}{2}
   \]

The final answer is \( \boxed{ a = \frac{b - \sqrt{b^2 - 9}}{2} } \)