In the tetrahedron $A B C D$, the lengths of two skew edges are $a$ and $b$, and the length of the segment connecting the midpoints of these edges is $k$. What is the maximum possible surface area and volume of the tetrahedron?
Let the endpoints of the edge of length $a$ be $A$ and $B$, and the other endpoints be $C$ and $D$, with the midpoints being $P$ and $Q$. We will also use the other notations in the diagram.

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Construct the parallelepiped circumscribed around the tetrahedron, i.e., the solid bounded by planes parallel to the pairs of opposite edges of the tetrahedron. One of the faces of this parallelepiped has the diagonal $A B = a$, and the other diagonal of this face is of length $b$. Therefore, the area of this face is $\frac{1}{2} a b \cdot \sin \varphi$. Let the volume of the tetrahedron be denoted by $V$.

It is known that the volume of the circumscribed parallelepiped is three times the volume of the tetrahedron. It is also evident that the height corresponding to the mentioned face of the parallelepiped, denoted by $m$, can be at most $k$, i.e., $m \leq k$. Equality holds here only if $P Q$ is perpendicular to the faces of the parallelepiped containing $A B$ and $C D$, and then $P Q$ is perpendicular to both $A B$ and $C D$. After this, we estimate the volume of the parallelepiped as follows:

$$
3 V=\frac{1}{2} a b \cdot(\sin \varphi) \cdot m \leq \frac{1}{2} \cdot a \cdot b \cdot k
$$

since $\sin \varphi \leq 1$, from which we get that $V \leq \frac{a \cdot b \cdot k}{6}$. Equality holds if and only if $\sin \varphi=1$ and $m=k$, which together are precisely satisfied if $A B$ and $C D$ are perpendicular to each other and to $P Q$. It is easy to see that such a tetrahedron exists.

For the surface area estimate, consider that the tetrahedron is bounded by two triangles on the edge $A B = a$ and two triangles on the edge $C D = b$. Since in a triangle, the height corresponding to a given side is not greater than the median to that side, the area of the triangle is at most $\frac{1}{2} \cdot a \cdot s_{a}$ with the usual notation.

Let the surface area of the tetrahedron be denoted by $A$. According to the above, we then have

$$
A \leq \frac{a(x+y)}{2}+\frac{b(r+s)}{2}
$$

where $x, y, r, s$ are the medians of the corresponding tetrahedron faces. Applying the cosine rule to the triangles $P Q D$ and $P Q C$, we get:

$$
x^{2}=k^{2}+\frac{b^{2}}{4}-k b \cdot \cos \alpha, \quad y^{2}=k^{2}+\frac{b^{2}}{4}+k b \cdot \cos \alpha
$$

where $\alpha$ is the angle $\angle D Q P$. Adding the two equations, we obtain

$$
x^{2}+y^{2}=2\left(k^{2}+\frac{b^{2}}{4}\right)
$$

By the arithmetic and quadratic mean inequality,

$$
\frac{x+y}{2} \leq \sqrt{\frac{x^{2}+y^{2}}{2}}
$$

thus, using (1),

$$
\frac{x+y}{2} \leq \sqrt{k^{2}+\frac{b^{2}}{4}}
$$

and similarly,

$$
\frac{r+s}{2} \leq \sqrt{k^{2}+\frac{a^{2}}{4}}
$$

Therefore,

$$
A \leq a \sqrt{k^{2}+\frac{b^{2}}{4}}+b \sqrt{k^{2}+\frac{a^{2}}{4}}
$$

Equality holds if and only if the segments $x, y, r, s$ are also heights in the corresponding triangles, and if $x=y$ and $r=s$. This means that $A B$ and $C D$ are perpendicular to each other and to $P Q$, i.e., the surface area and volume of the tetrahedron are simultaneously maximal.