Both facts follow from the lemma below.

Lemma. Consider a triangle some altitude of which has length at most $1$. If the triangle has an altitude of length at least $\sqrt{2}$, say the $X$-altitude, then the internal angular span at $X$ does not exceed $45^\circ$; equality holds if and only if the $X$-altitude has length $\sqrt{2}$, and the triangle is isosceles right with apex at one of the other two vertices.

To prove the lemma, let the $X$-altitude have length $h_X \ge \sqrt{2}$. Since the triangle has an altitude of length at most $1$, say $h_Y \le 1$, the internal angle at $X$ is not the largest angle of the triangle, so it does not exceed a right angle. And since $\sin X = (h_Y/h_X) \cdot \sin Y \le h_Y/h_X \le 1/\sqrt{2}$, the internal angular span at $X$ does not exceed $45^\circ$.

If the internal angular span at $X$ is $45^\circ$, then $1/\sqrt{2} = \sin X = (h_Y/h_X) \cdot \sin Y \le h_Y/h_X \le 1/\sqrt{2}$, forcing $h_X = \sqrt{2}$, $h_Y = 1$ and $\sin Y = 1$, so the triangle is indeed isosceles right with apex at $Y$. The converse is clear.

Back to the problem, begin by noticing that the four points may and will be assumed to be in strictly convex position; i.e., no point lies in the convex hull of the other three. Further, let $\text{dist}(X, YZ)$ denote the distance from the point $X$ to the line $YZ$, and notice that the four points may and will be labelled $A$, $B$, $C$, $D$, in circular order, so that $\text{dist}(A, BC) \ge \text{dist}(D, BC)$, $\text{dist}(A, CD) \ge \text{dist}(B, CD)$, $\text{dist}(B, AD) \ge \text{dist}(C, AD)$ and $\text{dist}(D, AB) \ge \text{dist}(C, AB)$.

a) It is clearly sufficient to show that one of the distances $\text{dist}(A, BC)$, $\text{dist}(A, CD)$, $\text{dist}(B, AD)$, $\text{dist}(D, AB)$ does not exceed $\sqrt{2}$. Suppose, if possible, this is not the case, notice that each of the triangles $ABC$, $ACD$, $ABD$ has an altitude of length at most $1$, and refer to the lemma to infer that the angular span of each of the angles $BAC$, $CAD$, $ABD$, $ADB$ is less than $45^\circ$, so the internal angular spans at the vertices of the triangle $ABD$ add up to less than $180^\circ$ — a contradiction.

b) Now each of the four distances $\text{dist}(A, BC)$, $\text{dist}(A, CD)$, $\text{dist}(B, AD)$, $\text{dist}(D, AB)$ is at least $\sqrt{2}$, and the lemma applies again to each of the triangles $ABC$, $ACD$, $ABD$, to deduce that the angular span of each of the angles $BAC$, $CAD$, $ABD$, $ADB$ does not exceed $45^\circ$. If one of the four distances above were greater than $\sqrt{2}$, then the angular span of the corresponding angle would be less than $45^\circ$ and the internal angular spans at the vertices of the triangle $ABD$ would again add up to less than $180^\circ$. Consequently, the four distances under consideration are all equal to $\sqrt{2}$, the corresponding angular spans are all $45^\circ$, and the conclusion follows from the equality case in the lemma.