Let $P(x)=x^2+bx+c$. Suppose $P(P(1))=P(P(-2))=0$ and $P(1)\neq P(-2)$. Then $P(0)=$
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[*] $-\frac{5}{2}$


[*] $-\frac{3}{2}$


[*] $-\frac{7}{4}$


[*] $\frac{6}{7}$
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1. Given the polynomial \( P(x) = x^2 + bx + c \), we know that \( P(P(1)) = 0 \) and \( P(P(-2)) = 0 \). This implies that \( P(1) \) and \( P(-2) \) are roots of the polynomial \( P(x) \).
2. We are given that \( P(1) \neq P(-2) \). Let's calculate \( P(1) \) and \( P(-2) \):
   \[
   P(1) = 1^2 + b \cdot 1 + c = 1 + b + c
   \]
   \[
   P(-2) = (-2)^2 + b \cdot (-2) + c = 4 - 2b + c
   \]
3. Since \( P(1) \) and \( P(-2) \) are roots of \( P(x) \), we can write:
   \[
   P(x) = (x - P(1))(x - P(-2))
   \]
4. By Vieta's formulas, the sum of the roots of \( P(x) \) is equal to \(-b\). Therefore:
   \[
   P(1) + P(-2) = -b
   \]
   Substituting the expressions for \( P(1) \) and \( P(-2) \):
   \[
   (1 + b + c) + (4 - 2b + c) = -b
   \]
5. Simplifying the equation:
   \[
   1 + b + c + 4 - 2b + c = -b
   \]
   \[
   5 - b + 2c = -b
   \]
   \[
   5 + 2c = 0
   \]
   \[
   2c = -5
   \]
   \[
   c = -\frac{5}{2}
   \]
6. Therefore, \( P(0) = c = -\frac{5}{2} \).

The final answer is \(\boxed{-\frac{5}{2}}\).