2.8. At the base of the pyramid lies a square. Two lateral faces are perpendicular to the base plane, while the other two are inclined to it at an angle of $45^{\circ}$. The middle-sized lateral edge is equal to $l$. Find the volume and the total surface area of the pyramid.
2.8. By the condition, $S C=l, \angle S B C=90^{\circ}$, $\angle S C B=45^{\circ}$ (Fig. 2.8). Hence, $S B=B C=\frac{l}{\sqrt{2}}$. We find $V=\frac{1}{3} B C^{2} \cdot S B=\frac{l^{3} \sqrt{2}}{12}$. The total surface area can be expressed as: $S_{\text {full }}=S_{\text {base }}+2 S_{\triangle S A B}+2 S_{\triangle S A D}$, since $S_{\triangle S A B}=S_{\triangle S B C}, S_{\triangle S A D}=S_{\triangle S C D}$. But

$$
S_{\triangle S A B}=\frac{1}{2}\left(\frac{l}{\sqrt{2}}\right)^{2}=\frac{l^{2}}{4}, S_{\triangle S A D}=\frac{1}{2} A D \cdot S A=\frac{1}{2} \cdot \frac{l}{\sqrt{2}} \cdot l=\frac{l^{2} \sqrt{2}}{4}
$$

Thus,

$$
S_{\text {full }}=\frac{l^{2}}{2}+\frac{l^{2}}{2}+\frac{l^{2} \sqrt{2}}{2}=\frac{l^{2}(2+\sqrt{2})}{2} \text {. }
$$

Answer: $\frac{l^{3} \sqrt{2}}{12} ; 0.5 l^{2}(2+\sqrt{2})$.