Let $l,\ m$ be the tangent lines passing through the point $A(a,\ a-1)$ on the line $y=x-1$ and touch the parabola $y=x^2$.
Note that the slope of $l$ is greater than that of $m$.

(1) Exress the slope of $l$ in terms of $a$.

(2) Denote $P,\ Q$ be the points of tangency of the lines $l,\ m$ and the parabola $y=x^2$.
Find the minimum area of the part bounded by the line segment $PQ$ and the parabola $y=x^2$.

(3) Find the minimum distance between the parabola $y=x^2$ and the line $y=x-1$.
### Part (1)
To find the slope of the tangent line \( l \) in terms of \( a \):

1. The equation of the tangent line to the parabola \( y = x^2 \) at a point \( (x_0, y_0) \) is given by:
   \[
   y = 2x_0 x - x_0^2
   \]
   Here, \( y_0 = x_0^2 \).

2. Since the tangent line passes through the point \( A(a, a-1) \), we substitute \( x = a \) and \( y = a-1 \) into the tangent line equation:
   \[
   a - 1 = 2x_0 a - x_0^2
   \]

3. Rearrange the equation to solve for \( x_0 \):
   \[
   x_0^2 - 2a x_0 + (a - 1) = 0
   \]

4. Solve the quadratic equation for \( x_0 \):
   \[
   x_0 = \frac{2a \pm \sqrt{4a^2 - 4(a-1)}}{2} = a \pm \sqrt{a^2 - a + 1}
   \]

5. The slopes of the tangent lines are given by \( k = 2x_0 \). Therefore, the slopes are:
   \[
   k_l = 2(a + \sqrt{a^2 - a + 1})
   \]
   \[
   k_m = 2(a - \sqrt{a^2 - a + 1})
   \]

Since the slope of \( l \) is greater than that of \( m \), we have:
\[
k_l = 2(a + \sqrt{a^2 - a + 1})
\]

### Part (2)
To find the minimum area of the part bounded by the line segment \( PQ \) and the parabola \( y = x^2 \):

1. The points of tangency \( P \) and \( Q \) are \( (x_1, x_1^2) \) and \( (x_2, x_2^2) \) where:
   \[
   x_1 = a - \sqrt{a^2 - a + 1}
   \]
   \[
   x_2 = a + \sqrt{a^2 - a + 1}
   \]

2. The equation of the line segment \( PQ \) is:
   \[
   y = 2ax - (a^2 - a + 1)
   \]

3. The area \( S \) between the parabola and the line segment \( PQ \) is given by:
   \[
   S = \int_{x_1}^{x_2} \left(2ax - (a^2 - a + 1) - x^2\right) \, dx
   \]

4. Evaluate the integral:
   \[
   S = \int_{x_1}^{x_2} (2ax - x^2 - (a^2 - a + 1)) \, dx
   \]

5. The roots \( x_1 \) and \( x_2 \) are symmetric about \( x = a \), so:
   \[
   x_1 + x_2 = 2a
   \]
   \[
   x_1 x_2 = a^2 - a + 1
   \]

6. The integral becomes:
   \[
   S = \left[ ax^2 - \frac{x^3}{3} - (a^2 - a + 1)x \right]_{x_1}^{x_2}
   \]

7. Substitute \( x_1 \) and \( x_2 \):
   \[
   S = \left[ a(x_2^2 - x_1^2) - \frac{1}{3}(x_2^3 - x_1^3) - (a^2 - a + 1)(x_2 - x_1) \right]
   \]

8. Simplify using the symmetry and properties of \( x_1 \) and \( x_2 \):
   \[
   S = \frac{4}{3}(a^2 - a + 1)^{3/2}
   \]

9. The minimum area occurs when \( a = \frac{1}{2} \):
   \[
   S_{\text{min}} = \frac{\sqrt{3}}{2}
   \]

### Part (3)
To find the minimum distance between the parabola \( y = x^2 \) and the line \( y = x - 1 \):

1. Let \( P(t, t^2) \) be a point on the parabola. The distance from \( P \) to the line \( y = x - 1 \) is given by:
   \[
   d = \frac{|t^2 - t + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|t^2 - t + 1|}{\sqrt{2}}
   \]

2. Minimize the distance function:
   \[
   d = \frac{|t^2 - t + 1|}{\sqrt{2}}
   \]

3. The minimum value of \( |t^2 - t + 1| \) occurs when \( t = \frac{1}{2} \):
   \[
   t^2 - t + 1 = \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}
   \]

4. Therefore, the minimum distance is:
   \[
   d_{\text{min}} = \frac{\frac{3}{4}}{\sqrt{2}} = \frac{3}{8} \sqrt{2}
   \]

The final answer is \( \boxed{\frac{3}{8} \sqrt{2}} \)