y=x^2, dy/dx=2x; y=-x^2+2x-5, dy/dx=-2x+2. These are the tangents at (x,y) for each equation, to find the tangent lines, you need to know at which point the tangent line is required, because it varies from point to point so I'm guessing that since you want two tangent lines, the two tangent lines touch each graph at different points. So that means we have two points on each tangent line to find.
We know the equation of a straight line has the form y=mx+b where m=dy/dx. If one tangent line is to touch both curves, then the tangent of one must be equal to the tangent of the other. So if we call the point on the first curve (x1,y1) and on the other (x2,y2), m=2x1=-2x2+2, and x1=-x2+1=m/2. The equation of the tangent line is y=2x1x+b=(-2x2+2)x+b. But the points (x1,x1^2) and (x2,-x2^2+2x2-5) must both lie on this line.
Therefore, x1^2=2x1^2+b and -x2^2+2x2-5=2x1x2+b. So b=-x1^2 from the first equation.
Substituting for b: -x2^2+2x2-5=2x1x2-x1^2. And x1^2-2x1x2-x2^2+2x2-5=0. We know x2=1-x1 so we can substitute for x2 and find x1. x1^2-2x1(1-x1)-(1-x1)^2+2(1-x1)-5=0; x1^2-2x1+2x1^2-1+2x1-x1^2+2-2x1-5=0.
2x1^2-2x1-4=0=x1^2-x1-2=(x1-2)(x1+1), so x1=2 or -1. Therefore, x2=1-x1=-1 or 2. y1=4 or 1, y2=-8 or -5. The four points are (2,4) and (-1,-8), and (-1,1) and (2,-5), a pair of points on each line. Also, b=-4 or -1 and m=4 or -2.
The two lines are y=4x-4 and y=-2x-1.