Find polynomials $f(x)$, $g(x)$, and $h(x)$, if they exist, such that for all $x$, \[|f(x)|-|g(x)|+h(x)=\begin{cases}-1 & \text{if }x<-1\\3x+2 &\text{if }-1\leq x\leq 0\\-2x+2 & \text{if }x>0.\end{cases}\]
1. We are given the piecewise function:
   \[
   |f(x)| - |g(x)| + h(x) = 
   \begin{cases}
   -1 & \text{if } x < -1 \\
   2 & \text{if } -1 \leq x \leq 0 \\
   -2x + 2 & \text{if } x > 0
   \end{cases}
   \]
   We need to find polynomials \( f(x) \), \( g(x) \), and \( h(x) \) that satisfy this equation for all \( x \).

2. Since \( x = -1 \) and \( x = 0 \) are the critical points where the behavior of the function changes, we can assume the general form of the function involving absolute values:
   \[
   F(x) = a|x+1| + b|x| + cx + d
   \]
   We need to match this form to the given piecewise function.

3. Consider the different cases for \( x \):

   - **Case 1: \( x < -1 \)**
     \[
     |x+1| = -(x+1) \quad \text{and} \quad |x| = -x
     \]
     Therefore,
     \[
     F(x) = a(-x-1) + b(-x) + cx + d = -ax - a - bx + cx + d = (c-a-b)x + (d-a)
     \]
     Given \( F(x) = -1 \) for \( x < -1 \), we have:
     \[
     (c-a-b)x + (d-a) = -1
     \]
     Since this must hold for all \( x < -1 \), the coefficient of \( x \) must be zero, and the constant term must be \(-1\):
     \[
     c - a - b = 0 \quad \text{and} \quad d - a = -1
     \]

   - **Case 2: \( -1 \leq x \leq 0 \)**
     \[
     |x+1| = x+1 \quad \text{and} \quad |x| = -x
     \]
     Therefore,
     \[
     F(x) = a(x+1) + b(-x) + cx + d = ax + a - bx + cx + d = (a-b+c)x + (a+d)
     \]
     Given \( F(x) = 2 \) for \( -1 \leq x \leq 0 \), we have:
     \[
     (a-b+c)x + (a+d) = 2
     \]
     Since this must hold for all \( -1 \leq x \leq 0 \), the coefficient of \( x \) must be zero, and the constant term must be \( 2 \):
     \[
     a - b + c = 0 \quad \text{and} \quad a + d = 2
     \]

   - **Case 3: \( x > 0 \)**
     \[
     |x+1| = x+1 \quad \text{and} \quad |x| = x
     \]
     Therefore,
     \[
     F(x) = a(x+1) + b(x) + cx + d = ax + a + bx + cx + d = (a+b+c)x + (a+d)
     \]
     Given \( F(x) = -2x + 2 \) for \( x > 0 \), we have:
     \[
     (a+b+c)x + (a+d) = -2x + 2
     \]
     Since this must hold for all \( x > 0 \), the coefficient of \( x \) must be \(-2\), and the constant term must be \( 2 \):
     \[
     a + b + c = -2 \quad \text{and} \quad a + d = 2
     \]

4. Solving the system of equations:
   \[
   \begin{cases}
   c - a - b = 0 \\
   a - b + c = 0 \\
   a + b + c = -2 \\
   a + d = 2
   \end{cases}
   \]

   From \( c - a - b = 0 \):
   \[
   c = a + b
   \]

   Substitute \( c = a + b \) into \( a - b + c = 0 \):
   \[
   a - b + (a + b) = 0 \implies 2a = 0 \implies a = 0
   \]

   Substitute \( a = 0 \) into \( c = a + b \):
   \[
   c = b
   \]

   Substitute \( a = 0 \) and \( c = b \) into \( a + b + c = -2 \):
   \[
   0 + b + b = -2 \implies 2b = -2 \implies b = -1
   \]

   Therefore, \( c = b = -1 \).

   Finally, substitute \( a = 0 \) into \( a + d = 2 \):
   \[
   0 + d = 2 \implies d = 2
   \]

5. Thus, the polynomials are:
   \[
   f(x) = 0, \quad g(x) = -x, \quad h(x) = -x + 2
   \]

The final answer is \( \boxed{ f(x) = 0, g(x) = -x, h(x) = -x + 2 } \).