Does there exist regular planar pentagon? How to prove or disprove that the boundary of any convex body in $\mathbb{R}^3$ (treated as a surface) includes 5 points which form a regular planar pentagon?

This is about a special case raised by Han de Bruijn: yes, you can inscribe a regular planar pentagon in a regular tetrahedron.
And you can do that in many ways, probably. Here is one of them.
Take a regular tetrahedron $ABCD$ with the edge length equal $1$. 
Let $E$ be a point on the $BC$ edge at distance $x$ from $B$. Similary $F$ on $DC$, $x$ apart from $D$.
The triangle $\triangle EFA$ is isosceles, degenerating continuously from the equilateral $\triangle BDA$ to the segment $CA$ as $x$ changes from $0$ to $1$.
That means $\epsilon = \angle EAF$ changes from $60^\circ$ to $0$. 'Somewhere' during the change it is $\epsilon = 36^\circ$. Where?
Let $y=AE=AF$ and $z=EF$. Due to the law of cosines in $\triangle ABE$ $$y^2 = 1^2 + x^2 - 2\cdot 1\cdot x\,\cos(\angle ABE) = x^2 + x + 1$$ and by the similarity $\triangle EFC \sim \triangle BDC$ $$EF = EC\times BD:BC$$ $$z = 1-x$$
In the isosceles $\triangle EFA$
$$\sin\frac\epsilon 2 = \frac z2:y$$
We want $\epsilon=36^\circ$:
$$\left .\sin 18^\circ = \frac{1-x}{\sqrt{x^2+x+1}} \ \ \right\vert \big(\big)^2$$
$$\sin^2 18^\circ = \frac{x^2-2x+1}{x^2+x+1} = 1 - \frac{3x}{x^2+x+1}$$
$$\cos^2 18^\circ = \frac{3x}{x^2+x+1}$$
$$x^2+x+1 = \frac{3x}{\cos^2 18^\circ}$$
$$x^2+\left(1-\frac{3}{\cos^2 18^\circ}\right)x+1 = 0$$
The parenthesized expression is approx. $-2.317$ and one of the roots falls between $0$ and $1$:
$$x\approx 0.573$$
Once we have $36^\circ$ at $A$ we can fit a regular pentagon $JKLMN$ in the $\triangle EFA$ so that $JK \subset FA$, $LM \subset AE$, $KL \parallel EF$. If it is too small or too big, scale it so that $N$ becomes the midpoint of $EF$.
