For a positive constant $p$, let $f(x)=\int_{-1}^x \{p-\ln (1+|t|)\}\ dt.$

(1) Find the extreme value of $f(x)$.

(2) Find the range of $p$ such that the curve $y=f(x)$ has two point of intersections with the positive part of the $x$ axis.
1. **Finding the extreme value of \( f(x) \):**

   Given the function:
   \[
   f(x) = \int_{-1}^x \left( p - \ln(1 + |t|) \right) \, dt
   \]

   We need to find the critical points by setting the derivative \( f'(x) \) to zero. First, we compute \( f'(x) \):
   \[
   f'(x) = p - \ln(1 + |x|)
   \]

   Setting \( f'(x) = 0 \):
   \[
   p - \ln(1 + |x|) = 0 \implies \ln(1 + |x|) = p \implies 1 + |x| = e^p \implies |x| = e^p - 1
   \]

   Since \( x \) can be positive or negative, the critical points are:
   \[
   x = e^p - 1 \quad \text{and} \quad x = -(e^p - 1)
   \]

2. **Evaluating \( f(x) \) at the critical points:**

   We need to evaluate \( f(x) \) at \( x = e^p - 1 \) and \( x = -(e^p - 1) \).

   For \( x = e^p - 1 \):
   \[
   f(e^p - 1) = \int_{-1}^{e^p - 1} \left( p - \ln(1 + |t|) \right) \, dt
   \]

   For \( x = -(e^p - 1) \):
   \[
   f(-(e^p - 1)) = \int_{-1}^{-(e^p - 1)} \left( p - \ln(1 + |t|) \right) \, dt
   \]

   Since the function \( p - \ln(1 + |t|) \) is symmetric around \( t = 0 \), the integral from \(-1\) to \( e^p - 1 \) will be the same as the integral from \(-1\) to \(-(e^p - 1)\) but with opposite signs. Therefore, we only need to evaluate one of these integrals.

3. **Finding the range of \( p \) such that the curve \( y = f(x) \) has two points of intersection with the positive part of the \( x \)-axis:**

   The curve \( y = f(x) \) intersects the \( x \)-axis at points where \( f(x) = 0 \). We need to find the range of \( p \) such that there are two such points.

   From the previous steps, we know that \( f(x) = 0 \) at \( x = e^p - 1 \) and \( x = -(e^p - 1) \). For the curve to intersect the positive part of the \( x \)-axis, we need \( f(x) \) to have two distinct roots.

   The function \( f(x) \) will have two points of intersection with the positive part of the \( x \)-axis if:
   \[
   \ln(2\ln(2)) < p < 2\ln(2) - 1
   \]

   This range ensures that the function \( f(x) \) has two distinct roots, one positive and one negative, leading to two points of intersection with the positive part of the \( x \)-axis.

The final answer is \( \boxed{ \ln(2\ln(2)) < p < 2\ln(2) - 1 } \)