Tangents are drawn to circle `x^(2)+y^(2)=1` at its iontersection points (distinct) with the circle `x^(2)+y^(2)+(lambda-3)x+(2lambda+2)y+2=0`. The locus of intersection of tangents is a straight line, then the slope of that straight line is .

Answer

To solve the problem, we need to find the slope of the straight line that represents the locus of the intersection of tangents drawn to the circle \( x^2 + y^2 = 1 \) at its intersection points with the second circle given by \( x^2 + y^2 + (\lambda - 3)x + (2\lambda + 2)y + 2 = 0 \).

### Step-by-Step Solution:

1. **Identify the circles**:
- The first circle is \( C_1: x^2 + y^2 = 1 \).
- The second circle is \( C_2: x^2 + y^2 + (\lambda - 3)x + (2\lambda + 2)y + 2 = 0 \).

2. **Find the intersection points**:
- To find the intersection points of the circles, we can set the equations equal to each other. We can rewrite \( C_2 \) as:
\[ x^2 + y^2 + (\lambda - 3)x + (2\lambda + 2)y + 2 = 0 \]
- Substituting \( x^2 + y^2 = 1 \) into the equation of \( C_2 \):
\[ 1 + (\lambda - 3)x + (2\lambda + 2)y + 2 = 0 \]
\[ (\lambda - 3)x + (2\lambda + 2)y + 3 = 0 \]

3. **Equation of the chord of intersection**:
- The equation of the chord of intersection can be expressed as:
\[ S_1 - S_2 = 0 \]
- Where \( S_1 = x^2 + y^2 - 1 \) and \( S_2 = x^2 + y^2 + (\lambda - 3)x + (2\lambda + 2)y + 2 \).
- Thus, we have:
\[ (x^2 + y^2 - 1) - (x^2 + y^2 + (\lambda - 3)x + (2\lambda + 2)y + 2) = 0 \]
- Simplifying gives:
\[ -(\lambda - 3)x - (2\lambda + 2)y - 3 = 0 \]
- Rearranging:
\[ (\lambda - 3)x + (2\lambda + 2)y = -3 \]

4. **Finding the slope of the line**:
- The equation can be rearranged to the form \( Ax + By + C = 0 \):
\[ (\lambda - 3)x + (2\lambda + 2)y + 3 = 0 \]
- The slope \( m \) of the line is given by:
\[ m = -\frac{A}{B} = -\frac{\lambda - 3}{2\lambda + 2} \]

5. **Condition for the locus to be a straight line**:
- For the locus of intersection of the tangents to be a straight line, the coefficients of \( \lambda \) must be constant. This means that the expression must yield a linear relationship in \( x \) and \( y \).

6. **Finding the specific slope**:
- We can set \( \lambda = 3 - 3h \) and \( k = 2\lambda + 2 \) to find the slope:
\[ 6h - 3k = 8 \]
- Substituting \( h \) and \( k \) gives us:
\[ 6x - 3y = 8 \]
- Rearranging gives:
\[ y = 2x - \frac{8}{3} \]
- Thus, the slope \( m = 2 \).

### Final Answer:
The slope of the straight line is \( \boxed{2} \).