Given a function $f(x)=x^{2}+ax+b$ with $a,b \in R$, if there exists a real number $m$ such that $\left| f(m) \right| \leq \frac{1}{4}$ and $\left| f(m+1) \right| \leq \frac{1}{4}$, then find the maximum and minimum of the value of $\Delta=a^{2}-4b$.
Given the function \( f(x) = x^2 + ax + b \) with \( a, b \in \mathbb{R} \), and the conditions \( \left| f(m) \right| \leq \frac{1}{4} \) and \( \left| f(m+1) \right| \leq \frac{1}{4} \) for some real number \( m \), we need to find the maximum and minimum values of \( \Delta = a^2 - 4b \).

1. **Express \( f(x) \) in a different form:**
   \[
   f(x) = x^2 + ax + b = \frac{(2x + a)^2 - a^2 + 4b}{4}
   \]
   Therefore,
   \[
   f(x) = \frac{(2x + a)^2 - \Delta}{4}
   \]
   where \( \Delta = a^2 - 4b \).

2. **Substitute \( x = m \) and \( x = m+1 \) into the function:**
   \[
   f(m) = \frac{(2m + a)^2 - \Delta}{4}
   \]
   \[
   f(m+1) = \frac{(2(m+1) + a)^2 - \Delta}{4} = \frac{(2m + 2 + a)^2 - \Delta}{4}
   \]

3. **Apply the given conditions:**
   \[
   \left| \frac{(2m + a)^2 - \Delta}{4} \right| \leq \frac{1}{4}
   \]
   \[
   \left| \frac{(2m + 2 + a)^2 - \Delta}{4} \right| \leq \frac{1}{4}
   \]

4. **Simplify the inequalities:**
   \[
   \left| (2m + a)^2 - \Delta \right| \leq 1
   \]
   \[
   \left| (2m + 2 + a)^2 - \Delta \right| \leq 1
   \]

5. **Let \( y = 2m + a \), then the inequalities become:**
   \[
   \left| y^2 - \Delta \right| \leq 1
   \]
   \[
   \left| (y + 2)^2 - \Delta \right| \leq 1
   \]

6. **Analyze the inequalities:**
   \[
   -1 \leq y^2 - \Delta \leq 1
   \]
   \[
   -1 \leq (y + 2)^2 - \Delta \leq 1
   \]

7. **Combine the inequalities:**
   \[
   y^2 - 1 \leq \Delta \leq y^2 + 1
   \]
   \[
   (y + 2)^2 - 1 \leq \Delta \leq (y + 2)^2 + 1
   \]

8. **Find the intersection of the ranges:**
   \[
   \max(y^2 - 1, (y + 2)^2 - 1) \leq \Delta \leq \min(y^2 + 1, (y + 2)^2 + 1)
   \]

9. **Evaluate the expressions:**
   \[
   y^2 - 1 \quad \text{and} \quad (y + 2)^2 - 1
   \]
   \[
   y^2 + 1 \quad \text{and} \quad (y + 2)^2 + 1
   \]

   Since \( (y + 2)^2 = y^2 + 4y + 4 \):
   \[
   (y + 2)^2 - 1 = y^2 + 4y + 3
   \]
   \[
   (y + 2)^2 + 1 = y^2 + 4y + 5
   \]

10. **Combine the results:**
    \[
    \max(y^2 - 1, y^2 + 4y + 3) \leq \Delta \leq \min(y^2 + 1, y^2 + 4y + 5)
    \]

    Since \( y^2 + 4y + 3 \geq y^2 - 1 \) and \( y^2 + 4y + 5 \geq y^2 + 1 \):
    \[
    y^2 + 4y + 3 \leq \Delta \leq y^2 + 1
    \]

    This implies:
    \[
    0 \leq \Delta \leq 2
    \]

Conclusion:
\[
\boxed{0 \leq \Delta \leq 2}
\]