Take a tetrahedron $ABCD$ which has $k$ edges of measure $a$ and the others of length $1$. $\textbf{Switch (k)}\ \{$ $\textbf{Case 1:}$ WLOG $CD=a$ and the measure of the other edges is $1$. Then if $M$ is the midpoint of $AB$ we have $MC+MD>CD\iff \sqrt{3}> a$. So $a<\sqrt3$ is a necessary condition. It is easy to ascertain that this is also a sufficient condition for the existence of $ABCD$. $\textbf{Case 2:}$ We have two subcases: $\textbf{(a)}$ Two opposite edges measure $a$ and the other edges are of length $1$, say $AB=CD=a$. Since $ABCD$ is isosceles and its faces are isosceles triangles which are acute then $ABCD$ exists if and only if its faces exist, that is when $a<2$. $\textbf{(b)}$ Two edges emanating from the same vertex have length $a$ and the others have length $1$, say $AD=BD=a$. Then $D$ lies on the circle $\omega$ with center $C$ which is perpendicular to the edge $AB$. Thus $AD=a$ attains its extremal values when $D$ attains its extremal values, that is when $D$ tends to one of the points of intersection of $\omega$ and the plane $ABC$. Hence we get $\sqrt{2-\sqrt3}<a<\sqrt{2+\sqrt3}$. Of course this is also a sufficient condition for the existence of $ABCD$. Thus we conclude that the desired necessary and sufficient condition for $k=2$ is $a\in (0, 2)\cup\left(\sqrt{2-\sqrt3}, \sqrt{2+\sqrt3}\right)=\left(0, \sqrt{2+\sqrt3}\right).$ $\textbf{Case 3:}$ We have three subcases: $\textbf{(a)}$ Three edges of length $a$ emanate from the same vertex, say $D$. Then $ABCD$ exists if and only if $a$ is greater than the circumradius of $\triangle ABC$, that is when $a>1/\sqrt3$. $\textbf{(b)}$ There is an open broken line with $3$ edges of length $a$ between the edges of $ABCD$. WLOG $AD=DB=BC=a$. Then $ABCD$ exists if and only if its faces exist which is just $1/2<a<2$. $\textbf{(c)}$ A face has all edges of length $a$. WLOG $AB=BC=CA=a$. This is the dual of the subcase $\textbf{(a)}$ so we get $a<\sqrt3$. Thus we conclude that for $k=3$ $ABCD$ exists for $a\in(1\sqrt3, +\infty)\cup(1/2, 2)\cup(0, \sqrt3)=(0, +\infty)$. $\textbf{Case 4:}$ This case is dual to the $\textbf{case 2}$ so the answer is $a\in\left(1/\sqrt{2+\sqrt3}, 1/\sqrt{2-\sqrt3}\right)\cup(1/2, +\infty)=\left(1/2, 1/\sqrt{2-\sqrt3}\right)$. $\textbf{Case 5:}$ This case is dual to the $\textbf{case 1}$ so the answer is $a\in (1/\sqrt3, +\infty)$. $\textbf{\}}$