1.  The Line $y=px+q$ intersects $y=x^2-x$, but not intersect $y=|x|+|x-1|+1$, then illustlate range of $(p,q)$ and find the area.
1. **Define the functions and conditions:**
   Let \( f(x) = x^2 - x \) and \( g(x) = |x| + |x-1| + 1 \). We need to find the range of \((p, q)\) such that the line \( y = px + q \) intersects \( y = x^2 - x \) but does not intersect \( y = |x| + |x-1| + 1 \).

2. **Intersection with \( y = x^2 - x \):**
   The line \( y = px + q \) intersects the parabola \( y = x^2 - x \) if the quadratic equation \( x^2 - (p+1)x + q = 0 \) has real roots. This requires the discriminant to be non-negative:
   \[
   \Delta = (p+1)^2 - 4q \geq 0
   \]
   Solving for \( q \), we get:
   \[
   q \leq \frac{(p+1)^2}{4}
   \]

3. **Non-intersection with \( y = |x| + |x-1| + 1 \):**
   The function \( g(x) = |x| + |x-1| + 1 \) can be broken into three linear segments:
   - For \( x \leq 0 \): \( g(x) = 1 - 2x \)
   - For \( 0 \leq x \leq 1 \): \( g(x) = 2 \)
   - For \( x \geq 1 \): \( g(x) = 2x - 1 \)

   The line \( y = px + q \) must lie below \( g(x) \) for all \( x \):
   - For \( x \leq 0 \): \( px + q < 1 - 2x \)
   - For \( 0 \leq x \leq 1 \): \( px + q < 2 \)
   - For \( x \geq 1 \): \( px + q < 2x - 1 \)

4. **Analyzing the inequalities:**
   - For \( x \leq 0 \): \( px + q < 1 - 2x \)
     \[
     q < 1 - (2 + p)x \quad \text{(since \( x \leq 0 \))}
     \]
     The most restrictive condition is when \( x = 0 \):
     \[
     q < 1
     \]

   - For \( 0 \leq x \leq 1 \): \( px + q < 2 \)
     \[
     q < 2 \quad \text{(since \( 0 \leq x \leq 1 \))}
     \]

   - For \( x \geq 1 \): \( px + q < 2x - 1 \)
     \[
     q < 2x - (p+1)x \quad \text{(since \( x \geq 1 \))}
     \]
     The most restrictive condition is when \( x = 1 \):
     \[
     q < 2 - p
     \]

5. **Combining the conditions:**
   - For \( p \leq 0 \):
     \[
     q < 2
     \]
   - For \( p > 0 \):
     \[
     q < 2 - p
     \]

6. **Range of \((p, q)\):**
   Combining the conditions from the discriminant and the non-intersection:
   \[
   q \leq \frac{(p+1)^2}{4} \quad \text{and} \quad q < \begin{cases} 2 & \text{if } p \leq 0 \\ 2 - p & \text{if } p > 0 \end{cases}
   \]

7. **Finding the area:**
   The area of the region in the \((p, q)\)-plane is bounded by these inequalities. We need to integrate over the valid region:
   \[
   \text{Area} = \int_{-2}^{0} \left(2 - \frac{(p+1)^2}{4}\right) dp + \int_{0}^{2} \left(2 - p - \frac{(p+1)^2}{4}\right) dp
   \]

   - For \( -2 \leq p \leq 0 \):
     \[
     \int_{-2}^{0} \left(2 - \frac{(p+1)^2}{4}\right) dp = \int_{-2}^{0} \left(2 - \frac{p^2 + 2p + 1}{4}\right) dp = \int_{-2}^{0} \left(2 - \frac{p^2}{4} - \frac{p}{2} - \frac{1}{4}\right) dp
     \]
     \[
     = \int_{-2}^{0} \left(\frac{7}{4} - \frac{p^2}{4} - \frac{p}{2}\right) dp
     \]
     \[
     = \left[\frac{7p}{4} - \frac{p^3}{12} - \frac{p^2}{4}\right]_{-2}^{0}
     \]
     \[
     = \left(0 - 0 - 0\right) - \left(\frac{7(-2)}{4} - \frac{(-2)^3}{12} - \frac{(-2)^2}{4}\right)
     \]
     \[
     = 0 - \left(-\frac{14}{4} + \frac{8}{12} + \frac{4}{4}\right)
     \]
     \[
     = \frac{14}{4} - \frac{2}{3} - 1
     \]
     \[
     = \frac{14}{4} - \frac{2}{3} - \frac{4}{4}
     \]
     \[
     = \frac{14 - 4}{4} - \frac{2}{3}
     \]
     \[
     = \frac{10}{4} - \frac{2}{3}
     \]
     \[
     = \frac{5}{2} - \frac{2}{3}
     \]
     \[
     = \frac{15}{6} - \frac{4}{6}
     \]
     \[
     = \frac{11}{6}
     \]

   - For \( 0 \leq p \leq 2 \):
     \[
     \int_{0}^{2} \left(2 - p - \frac{(p+1)^2}{4}\right) dp = \int_{0}^{2} \left(2 - p - \frac{p^2 + 2p + 1}{4}\right) dp = \int_{0}^{2} \left(2 - p - \frac{p^2}{4} - \frac{p}{2} - \frac{1}{4}\right) dp
     \]
     \[
     = \int_{0}^{2} \left(\frac{7}{4} - p - \frac{p^2}{4} - \frac{p}{2}\right) dp
     \]
     \[
     = \int_{0}^{2} \left(\frac{7}{4} - \frac{3p}{2} - \frac{p^2}{4}\right) dp
     \]
     \[
     = \left[\frac{7p}{4} - \frac{3p^2}{4} - \frac{p^3}{12}\right]_{0}^{2}
     \]
     \[
     = \left(\frac{7(2)}{4} - \frac{3(2)^2}{4} - \frac{(2)^3}{12}\right) - \left(0 - 0 - 0\right)
     \]
     \[
     = \left(\frac{14}{4} - \frac{12}{4} - \frac{8}{12}\right)
     \]
     \[
     = \left(\frac{14}{4} - \frac{12}{4} - \frac{2}{3}\right)
     \]
     \[
     = \left(\frac{14 - 12}{4} - \frac{2}{3}\right)
     \]
     \[
     = \left(\frac{2}{4} - \frac{2}{3}\right)
     \]
     \[
     = \left(\frac{1}{2} - \frac{2}{3}\right)
     \]
     \[
     = \left(\frac{3}{6} - \frac{4}{6}\right)
     \]
     \[
     = -\frac{1}{6}
     \]

   Adding the two areas:
   \[
   \text{Total Area} = \frac{11}{6} + \left(-\frac{1}{6}\right) = \frac{10}{6} = \frac{5}{3}
   \]

The final answer is \(\boxed{\frac{5}{3}}\).