Answer
(a) See explanations.
(b) $109.5^{\circ}$
Work Step by Step
(a) To show that the tetrahedron is regular, we need to prove that all the sides are of equal length. Since all the vertices are given, use the distance formula we have the side length $AB=\sqrt {1^2+1^2}=\sqrt 2$, we can show that other lengths are the same $AB=AC=AD=BC=BD=CD=\sqrt 2$, thus the tetrahedron is regular.
(b) Given the center at $E(\frac{1}{2},\frac{1}{2},\frac{1}{2})$, we can form two vectors $\vec{EA}=\langle \frac{1}{2}, -\frac{1}{2}. -\frac{1}{2} \rangle, |\vec{EA}|=\sqrt {3\times\frac{1}{4}}=\frac{\sqrt 3}{2}$ and $\vec{EB}=\langle -\frac{1}{2}, \frac{1}{2}. -\frac{1}{2} \rangle, |\vec{EB}|=\frac{\sqrt 3}{2}$.
Use the angle formula, we have $cos\angle AEB=\frac{\vec{EA}\cdot\vec{EB}}{|\vec{EA}||\vec{EB}|}=\frac{-1/4-1/4+1/4}{(\sqrt 3/2)^2}=-\frac{1}{4}\times\frac{4}{3}=-\frac{1}{3}$, thus $\theta=109.5^{\circ}$
