(a) Let $e, f$ be the diagonal lengths and $\theta$ be the angle between them. Then the sides of the inscribed parallelogram will be of lengths $\frac{\alpha}{\alpha+1}e$ and $\frac{1}{\alpha+1}f$ for some $\alpha>0$. Then we need to prove $\frac{1}{2}ef\sin\theta\ge 2\frac{\alpha}{(\alpha+1)^2}ef\sin\theta$ or equivalently $(\alpha-1)^2\ge 0$ which is obvious. (b) I think that the word "perpendicular" in the formulation should be replaced by "parallel". Project the tetrahedron's vertices on the secant plane. Then we get a convex quadrilateral which's doubled area is less than the surface area of the tetrahedron. Now simply apply (a).