The base of the pyramid is a parallelogram $ABCD$ with an area of $m^{2}$. It is known that $BD$ is perpendicular to $AD$. The dihedral angles at the edges $AD$ and $BC$ are $45^{\circ}$, and at the edges $AB$ and $CD-60^{\circ}$. Find the lateral surface area and volume of the pyramid.
Let $P$ be the vertex of the pyramid, $P H$ be its height. Since the dihedral angles at the edges $A D$ and $B C$ are equal, the point $H$ is equidistant from the lines $A D$ and $B C$, and therefore lies on the midline of the parallelogram $A B C D$ parallel to the sides $A D$ and $B C$. Similarly, we can prove that the point $H$ lies on the midline of the parallelogram $A B C D$ parallel to the other two sides. Since the midlines of the parallelogram intersect at its center, $H$ is the point of intersection of the diagonals of the parallelogram $A B C D$.

By the condition $H D \perp A D$. Then, by the theorem of three perpendiculars, $P D \perp A D$, so $H D P$ is the linear angle of the dihedral angle at the edge $A D$. Similarly, we can prove that $H B P$ is the linear angle of the dihedral angle at the edge $B C$. Therefore, $\angle H D P=\angle H B P=45^{\circ}$.

Let $P K H$ be the linear angle at the edge $A B$. By the condition $\angle P K H=60^{\circ}$. Triangle $A H B$ is the orthogonal projection of the face $A P B$ onto the base plane, so

$$
S(A P B)=S(A H B) / \cos \angle P K H=\frac{1}{4} m^{2} /\left(\frac{1}{2}\right)=\frac{1}{2} m^{2}
$$

Similarly, we find that

$$
\begin{gathered}
S(D P C)=\frac{1}{2} m^{2}, S(A P D)=S(A H D) / \cos 45^{\circ}=\frac{1}{4} m^{2} \sqrt{2} \\
S(B P C)=S(B H C) / \cos 45^{\circ}=\frac{1}{4} m^{2} \sqrt{2}
\end{gathered}
$$

Therefore,

$$
S(A P B)+S(D P C)+S(A P D)+S(B P C)=m^{2}+\frac{1}{2} m^{2} \sqrt{2}=m^{2}(1+\sqrt{2} / 2)
$$

Let $P H=x, \angle A B D=\alpha$. Then

$$
\begin{gathered}
B H=D H=P H=x, \quad K H=P H \cdot \operatorname{ctg} \angle P K H=x \cdot \operatorname{ctg} 60^{\circ}=\frac{1}{3} x \sqrt{3}, \\
\sin \alpha=\sin \angle A B D=K H / B H=\frac{1}{3} x \sqrt{3} / x=\frac{1}{3} \sqrt{3}, \cos \alpha=\sqrt{2 / 3}, \\
\operatorname{tg} \alpha=\sqrt{2} / 2, A D=B D \cdot \operatorname{tg} \alpha=2 x \cdot \sqrt{2} / 2=x \sqrt{2} .
\end{gathered}
$$

Then

$$
S(A B C D)=2 \cdot S(A B D)=A D \cdot B D=x \sqrt{2} \cdot 2 x=2 x^{2} \sqrt{2}=m^{2}
$$

from which $P H=x=m / \sqrt{2 \sqrt{2}}$.

Therefore,

$$
V(A B C D)=\frac{1}{3} S(A B C D) \cdot P H=\frac{1}{3} m^{2} \cdot m / \sqrt{2 \sqrt{2}}=\frac{1}{3} m^{3} / \sqrt{2 \sqrt{2}}=\frac{1}{6} m^{3} \sqrt[4]{2}
$$

## Answer

$$
m^{2}(1+1 / \sqrt{2}), m^{3} \cdot \sqrt[4]{2} / 6
$$