We are given a tetrahedron with two edges of length $a$ and the remaining four edges of length $b$ where $a$ and $b$ are positive real numbers. What is the range of possible values for the ratio $v=a/b$?
We are given a tetrahedron \(ABCD\) with two edges of length \(a\) and the remaining four edges of length \(b\). We need to find the range of possible values for the ratio \(v = \frac{a}{b}\).

1. **Case 1: The 4 congruent lengths form an equilateral triangle.**
   - Assume without loss of generality (WLOG) that \(AB = AC = BC = AD = b\).
   - Since \(DB = DC\), the locus of point \(D\) is a circle of radius \(b\), centered at \(A\), and perpendicular to the plane determined by \(\triangle ABC\).
   - The distance \(a\) can be calculated using the coordinates of \(D\) on this circle. The minimum and maximum distances from \(A\) to \(D\) on this circle are given by:
     \[
     a_{\text{min}} = \frac{\sqrt{6} - \sqrt{2}}{2}b \quad \text{and} \quad a_{\text{max}} = \frac{\sqrt{6} + \sqrt{2}}{2}b
     \]
   - Therefore, in this case, the ratio \(v = \frac{a}{b}\) ranges from \(\frac{\sqrt{6} - \sqrt{2}}{2}\) to \(\frac{\sqrt{6} + \sqrt{2}}{2}\).

2. **Case 2: The 4 congruent lengths do not form a triangle.**
   - Assume WLOG that \(AB = AC = DB = DC = b\).
   - In this configuration, \(BC\) and \(AD\) can be arbitrarily small, so there is no lower bound on \(a\) (except that it has to be positive).
   - However, if \(BC\) exceeds \(\sqrt{2}b\), then the distance from \(A\) to \(BC\) must be less than \(\frac{\sqrt{2}}{2}b\), so \(AD\) must be less than \(\sqrt{2}b\), which is a contradiction.
   - Therefore, the upper bound for \(a\) in this case is \(\sqrt{2}b\), which can only be attained when \(ABCD\) is a square.
   - Thus, in this case, the ratio \(v = \frac{a}{b}\) ranges from \(0\) to \(\sqrt{2}\).

3. **Combining the intervals:**
   - From Case 1, the interval for \(v\) is \(\left(\frac{\sqrt{6} - \sqrt{2}}{2}, \frac{\sqrt{6} + \sqrt{2}}{2}\right)\).
   - From Case 2, the interval for \(v\) is \((0, \sqrt{2})\).
   - The combined interval for \(v\) is the union of these two intervals, which is \(\left(0, \frac{\sqrt{6} + \sqrt{2}}{2}\right)\).

The final answer is \(\boxed{\left(0, \frac{\sqrt{6} + \sqrt{2}}{2}\right)}\).