a) First we find the values of $a$ for which the parabola is tangent to the hyperbola (see fig.). Let $\alpha$ be the abscissa of the tangency point. At this point the derivatives of functions $x^2-a$ and $1/x$ are equal, i.e. $2\alpha = -1/\alpha^2$, whence $\alpha = -\frac{1}{\sqrt[3]{2}}$ and the ordinate of the tangency point equals $\frac{1}{\alpha} = -\sqrt[3]{2}$. Then the parabola equation for this point gives $-\sqrt[3]{2} = \left(-\frac{1}{\sqrt[3]{2}}\right)^2 - a$, whence $a = \frac{3}{2}\sqrt[3]{2}$. It is clear that for $a < \frac{3}{2}\sqrt[3]{2}$ the parabola and the hyperbola have exactly one common point and for $a > \frac{3}{2}\sqrt[3]{2}$ they have exactly three common points.

b) Each intersection point of the parabola and the hyperbola satisfy the system of equations $y = x^2 - a$ and $xy = 1$. Hence it satisfy the equations $y^2 = x^2y - ay$ and $y^2 = x - ay$. Subtracting the latter equation from $y = x^2 - a$, we obtain the equation $y - y^2 = x^2 - a - x + ay$, which is equivalent to $(x - \frac{1}{2})^2 + (y - \frac{a-1}{2})^2 = \frac{(a+1)^2+1}{4}$.
This is the equation of the circle centered at $(\frac{1}{2}; -\frac{a-1}{2})$. Thus, the required locus is the vertical ray $x = \frac{1}{2}$ with the condition $y < -\frac{3}{4}\sqrt[3]{2} + \frac{1}{2}$.