a) Let $\mathcal{S}$ be the sphere tangent to the edges from the inside, and let $\mathcal{S}_X$ be the sphere tangent to the edges of the face opposite to vertex $X$, and the extensions of the edges from $X$. Consider say $\mathcal{S}$ and $\mathcal{S}_S$. If we cut these spheres by plane $ABC$, in both cases we get a circle on plane $ABC$ tangent to the segments $AB,BC,CA$ - this is the incircle of $\triangle ABC$. Hence $\mathcal{S}$ and $\mathcal{S}_S$ touches the edges $AB,BC,CA$ in the same points. Now cut the two spheres by plane $SAB$. We get the incircle and $S$-excircle of $\triangle SAB$. We have proven that $\mathcal{S}$ and $\mathcal{S}_S$ touch $AB$ in the same point, so the incircle and $S$-excircle of $\triangle SAB$ touch $AB$ at the same point. This implies that $SA=SB$. Similarly, we get that any two edges are the same length, therefore the tetrahedron is indeed regular. b) Look at the regular tetrahedron $SABC$. By symmetry, the center $O$ of the regular tetrahedron $SABC$ is of equal distance $d$ from all its edges, so taking $\mathcal{S}(O,d)$ works. Let the incenter of say $\triangle SAB$ be $I$, the $S$-excenter of $\triangle SAB$ be $J$. Consider the homothety with center $S$ that maps $I$ to $J$. This homothety maps the incircle $\omega$ of $\triangle SAB$ to the $S$-excircle of $\triangle SAB$; since $\omega\subset \mathcal{S}$, we have $\omega'\subset\mathcal{S}'$ and therefore $\mathcal{S}'$ also touches edge $AB$. By rotational symmetry, $\mathcal{S}'$ also touches edges $BC$ and $CA$. Moreover, because $\mathcal{S}$ touched edges $SA,SB,SC$, $\mathcal{S}'$ will touch the extensions of $SA,SB,SC$. This proves that $\mathcal{S}_S=\mathcal{S}'$ is up for the job. By symmetry, so do $\mathcal{S}_A,\mathcal{S}_B,\mathcal{S}_C$ exist.