Define a parabola $C$ by $y=x^2+1$ on the coordinate plane. Let $s,\ t$ be real numbers with $t<0$. Denote by $l_1,\ l_2$ the tangent lines drawn from the point $(s,\ t)$ to the parabola $C$.

(1) Find the equations of the tangents $l_1,\ l_2$.

(2) Let $a$ be positive real number. Find the pairs of $(s,\ t)$ such that the area of the region enclosed by $C,\ l_1,\ l_2$ is $a$.
1. **Finding the equations of the tangents \( l_1 \) and \( l_2 \):**

   Let the point of tangency on the parabola \( y = x^2 + 1 \) be \( (w, w^2 + 1) \). The equation of the tangent line to the parabola at this point is given by:
   \[
   y - (w^2 + 1) = 2w(x - w)
   \]
   Simplifying, we get:
   \[
   y = 2wx - w^2 + 1
   \]

   Since the point \( (s, t) \) lies on this tangent line, we substitute \( (s, t) \) into the equation:
   \[
   t = 2ws - w^2 + 1
   \]
   Rearranging, we get a quadratic equation in \( w \):
   \[
   w^2 - 2ws + (t - 1) = 0
   \]
   Solving this quadratic equation for \( w \), we use the quadratic formula:
   \[
   w = \frac{2s \pm \sqrt{4s^2 - 4(t - 1)}}{2} = s \pm \sqrt{s^2 + 1 - t}
   \]
   Let \( w_1 = s - \sqrt{s^2 + 1 - t} \) and \( w_2 = s + \sqrt{s^2 + 1 - t} \). The equations of the tangent lines \( l_1 \) and \( l_2 \) are:
   \[
   l_1: y = 2w_1x - w_1^2 + 1
   \]
   \[
   l_2: y = 2w_2x - w_2^2 + 1
   \]

2. **Finding the pairs \((s, t)\) such that the area of the region enclosed by \( C, l_1, l_2 \) is \( a \):**

   Let \( \alpha = \sqrt{s^2 + 1 - t} \). The points of tangency are \( w_1 = s - \alpha \) and \( w_2 = s + \alpha \).

   The area \( A(s, t) \) enclosed by the parabola and the tangent lines is given by:
   \[
   A(s, t) = \int_{w_1}^s \left( (x^2 + 1) - (2w_1x - w_1^2 + 1) \right) dx + \int_s^{w_2} \left( (x^2 + 1) - (2w_2x - w_2^2 + 1) \right) dx
   \]

   Simplifying the integrands:
   \[
   A(s, t) = \int_{w_1}^s \left( x^2 - 2w_1x + w_1^2 \right) dx + \int_s^{w_2} \left( x^2 - 2w_2x + w_2^2 \right) dx
   \]

   Evaluating the integrals:
   \[
   \int_{w_1}^s \left( x^2 - 2w_1x + w_1^2 \right) dx = \left[ \frac{x^3}{3} - w_1x^2 + w_1^2x \right]_{w_1}^s
   \]
   \[
   \int_s^{w_2} \left( x^2 - 2w_2x + w_2^2 \right) dx = \left[ \frac{x^3}{3} - w_2x^2 + w_2^2x \right]_s^{w_2}
   \]

   Substituting the limits:
   \[
   A(s, t) = \left( \frac{s^3}{3} - w_1s^2 + w_1^2s \right) - \left( \frac{w_1^3}{3} - w_1^3 + w_1^3 \right) + \left( \frac{w_2^3}{3} - w_2s^2 + w_2^2s \right) - \left( \frac{s^3}{3} - w_2s^2 + w_2^2s \right)
   \]

   Simplifying further:
   \[
   A(s, t) = \frac{w_2^3 - w_1^3}{3} - w_1(s^2 - w_1^2) + w_1^2(s - w_1) - w_2(w_2^2 - s^2) + w_2^2(w_2 - s)
   \]

   Using \( w_1 = s - \alpha \) and \( w_2 = s + \alpha \):
   \[
   A(s, t) = \frac{(s + \alpha)^3 - (s - \alpha)^3}{3} - (s - \alpha)(2\alpha - \alpha^2) + (s - \alpha)^2\alpha - (s + \alpha)(2\alpha + \alpha^2) + (s + \alpha)^2\alpha
   \]

   Simplifying the expression:
   \[
   A(s, t) = \frac{2\alpha}{3}(3s^2 + \alpha^2) - 4\alpha s - 2\alpha^3 + 2\alpha(s^2 + \alpha^2)
   \]

   \[
   A(s, t) = 4\alpha \left( s^2 - s + \frac{\alpha^2}{6} \right)
   \]

   Setting \( A(s, t) = a \):
   \[
   4\alpha \left( s^2 - s + \frac{\alpha^2}{6} \right) = a
   \]

   Solving for \( s \) and \( t \) in terms of \( a \) is complex and may not have a simple closed-form solution. However, the equation above provides the relationship needed to find the pairs \( (s, t) \) for a given area \( a \).

The final answer is \( \boxed{ 4\alpha \left( s^2 - s + \frac{\alpha^2}{6} \right) = a } \)