Call the tetrahedron $ABCD$. We do casework on the position of the $4$ congruent lengths Case 1: The $4$ congruent lengths from an equilateral triangle. Assume WLOG that $AB=AC=BC=AD=1$. Notice that since $DB=DC$, the locus of where $D$ can be is a circle of radius $1$, centered at $A$, and perpendicular to plane determined by $ABC$. It is not hard to see that $a$ can only range from $\frac{\sqrt{6}-\sqrt{2}}{2}$ to $\frac{\sqrt{6}+\sqrt{2}}{2}$ which are only attainable when $D$ is on the plane determined by $ABC$. Nonetheless, our interval of this case $\Big(\frac{\sqrt{6}-\sqrt{2}}{2}, \frac{\sqrt{6}+\sqrt{2}}{2}\Big)$ Case 2: The $4$ congruent lengths do not form a triangle. Assume WLOG that $AB=AC=DB=DC=1$. Then notice that $BC$ and $AD$ can be arbitrarily small, so there is no lower bound on $a$ (except that it has to be positive). However, notice that if $BC$ exceeds $\sqrt{2}$, then the distance from $A$ to $BC$ must be less that $\frac{\sqrt{2}}{2}$, so $AD$ must the less than $\sqrt{2}$, contradiction. So our upper bound is $\sqrt{2}$ which can only be attained when $ABCD$ is a square. Nonetheless, our interval for this case is $(0, \sqrt{2})$. Combining these, we get the desired interval $\boxed{\Big(0, \frac{\sqrt{6}+\sqrt{2}}{2}\Big)}$