The answer is yes to both questions. Start with a regular (all faces equilateral triangles) tetrahedron $P_1'P_2'P_3'P_4'$, where the answer is clearly yes to both questions. Then apply an affine transformation $A$ (a combination of stretching, shearing, rotating, and translation) to bring $P_1'P_2'P_3'P_4'$ to $P_1P_2P_3P_4$. For notation, let $A=T\circ L$, where $L$ is linear and $T$ is a translation by some vector $u$.
The transformation will have modified the volume of each of the sub-tetrahedra in the same way: multiplying it by $\det L$. So the sub-tetrahedra of $P_1P_2P_3P_4$ all have the same volume.
The relation that $\frac{P_1'+P_2'+P_3'+P_4'}{4}=O'$ will continue to hold once the transformation is applied to each of these points: $$P_i=AP_i'=LP_i'+u$$
So $$\begin{align}
\frac{P_1+P_2+P_3+P_4}{4}&=\frac{LP_1'+LP_2'+LP_3'+LP_4'+4u}{4}\\
&=L\left(\frac{P_1'+P_2'+P_3'+P_4'}{4}\right)+u\\
&=LO'+u\\
&=AO'\\
\frac{P_1+P_2+P_3+P_4}{4}&=O
\end{align}$$
As it happens, the "face-angles" surrounding a tetrahedral vertex are all you need to know to determine the dihedral angles at that vertex (and vice versa). What happens at the vertex doesn't care about the opposite face.
The connection is made through what's called the "law of cosines for spherical geometry", although that terminology makes the results seem more exotic and scary than they really are. They follow from straightforward (if slightly tedious) Euclidean trig or vector methods.
Given face-angles $\alpha := \angle BDC$, $\beta := \angle CDA$, $\gamma := \angle ADB$, and writing simply $A$, $B$, $C$ for dihedral angles along edges $DA$, $DB$, $DC$, the relations are
$$\begin{align}
\cos\alpha &= \cos\beta\cos\gamma+\sin\beta\sin\gamma\cos A \\
\cos\beta &= \cos\gamma\cos\alpha+\sin\gamma\sin\alpha\cos B \\
\cos\gamma &= \cos\alpha\cos\beta +\sin\alpha\sin\beta\cos C
\end{align}$$
As a sanity check: Consider a corner of a room where two walls and meet the floor. The floor need not be square; let's say there's a face-angle $\alpha$ between the two wall-floor edges. We'll assume the walls themselves are upright in the sense that they meet along an edge that makes right face-angles with the wall-floor edges (so, $\beta=\gamma=90^\circ$). Substituting this information into the above relations gives
$$\begin{align}
\cos\alpha &= \cos90^\circ\cos90^\circ+\sin90^\circ\sin90^\circ\cos A \;\quad\to\quad &\cos\alpha &= \cos A\\
\cos90^\circ&= \cos90^\circ\cos\alpha\;\;+\sin90^\circ\sin\alpha\;\;\,\cos B
\;\quad\to\quad &0 &= \cos B\\
\cos90^\circ&= \;\;\cos\alpha\cos90^\circ+\;\sin\alpha\;\sin90^\circ\;\cos C
\,\quad\to\quad &0&=\cos C
\end{align}$$
These confirm that the walls are upright in another sense, making right dihedral angles with the floor: $B=C=90^\circ$. Moreover, the dihedral angle between the walls exactly matches the face-angle on the floor: $A=\alpha$.
These relations have counterparts that switch the roles of face- and dihedral angles (and change one sign, so be careful!):
$$\begin{align}
\cos A &= -\cos B\cos C+\sin B\sin C\cos \alpha \\
\cos B &= -\cos C\cos A+\sin C\sin A\cos \beta \\
\cos C &= -\cos A\cos B+\sin A\sin B\cos \gamma
\end{align}$$
Note that substituting dihedral angle information from the corner of our room ($B=C=90^\circ$) would give us the face-angle information ($\beta=\gamma=90^\circ$, $\alpha=A$).
Importantly, these calculations require no knowledge of the shape of the rest of the room. In the tetrahedral context: edge-lengths are unnecessary; face-angles alone determine dihedral angles, and vice-versa.
(Of course, edge-lengths can help you find any missing face-angles you might need, but this isn't the scenario described in the question.)
Best Answer
Allow me first to rename your angles in a logical manner in the following way :
$AB=2 \sin(c/2), BC=2 \sin(a/2), CA= 2 \sin(b/2)$ where a,b and c are angles $BDC,BDC$ and $CDA$ respectively.
(for example I don't find notation $AB=2 \cos(a/2)$ very logical because if we have two letters $A$ and $B$, the third one must be the missing letter $c$).
The volume $V$ of tetrahedron $DABC$ is equal to
$$V=\frac16 \det M \ \ \ \text{where} \ \ \ M:=\pmatrix{|&|&| \\ \vec{DA}&\vec{DB}&\vec{DC}\\|&|&|}$$
Squaring, we get :
$$36 V^2=\det(M)^2=\det(M^T)\det(M)=\det(M^TM)$$
i.e., the determinant of all dot products in this way :
$$36 V^2 = =\begin{vmatrix}\vec{DA}.\vec{DA}& \vec{DA}.\vec{DB}&\vec{DA}.\vec{DC}\\ \vec{DB}.\vec{DA}& \vec{DB}.\vec{DB}&\vec{DB}.\vec{DC}\\ \vec{DC}.\vec{DA}& \vec{DC}.\vec{DB}&\vec{DC}.\vec{DC}\end{vmatrix}$$
$$36 V^2=\begin{vmatrix}1&\cos c&\cos b\\ \cos c & 1 & \cos a\\ \cos b & \cos a & 1\end{vmatrix}$$
$$36 V^2=1+2\cos a \cos b \cos c -(\cos^2a +\cos^2 b + \cos^2 c) \tag{1}$$
Besides, volume $V$ can be computed as :
$$V=\frac13 h \times \text{area(DBC)}=\frac13 h \times \frac12 1 \times 1 \sin a$$
Squaring the previous relationship, we get :
$$36 V^2=h^2 \sin^2 a \tag{2}$$
The value of altitude's length $h$ is obtained from the comparison of (1) and (2) :
$$h^2=\frac{1}{\sin^2 a}(1+2\cos a \cos b \cos c -(\cos^2a +\cos^2 b + \cos^2 c))$$