Consider the parabola y=x^2-2px+p+1.

If the real number p has the property a<p<b, then the vertex of y is above the line y=−12x+13.
What are a and b?

The vertex lies at (p,-p^2+p+1). So, we need

-p^2+p+1 > -12p+13
p^2+11p-12 < 0
-12 < p < 1

1<p<12

Therefore
a=1
b=12

To determine the values of a and b, we need to find the x-coordinate of the vertex of the parabola and check if it is greater than the x-coordinate of the point where the parabola intersects the line y = -12x + 13.

The x-coordinate of the vertex of a parabola of the form y = ax^2 + bx + c can be found using the formula x = -b / (2a). In this case, the equation of the parabola is y = x^2 - 2px + p + 1. Comparing the equation with the standard form, we can conclude that a = 1, b = -2p, and c = p + 1.

Using the formula for the x-coordinate of the vertex, we substitute the values of a and b into the formula:
x = -(-2p) / (2 * 1)
x = 2p / 2
x = p

Now we need to find the y-coordinate of the vertex, which can be obtained by substituting the x-coordinate (p) into the equation of the parabola:
y = p^2 - 2p * p + p + 1
y = p^2 - 2p^2 + p + 1
y = -p^2 + p + 1

To determine if the vertex is above the line y = -12x + 13, we compare the y-coordinate of the vertex (-p^2 + p + 1) with the y-coordinate of the point of intersection.

Setting the equation of the line equal to the equation of the parabola, we have:
-p^2 + p + 1 = -12p + 13

Rearranging the equation, we get:
p^2 - 13p + 12 = 0

Factoring the quadratic equation, we have:
(p - 1)(p - 12) = 0

This gives us two possible values for p: p = 1 and p = 12.

Now we need to determine the values of a and b based on the range a < p < b. Since a < p < b, we arrange the values in ascending order:
a = 1, b = 12

Therefore, the values of a and b, satisfying the given conditions, are a = 1 and b = 12.