5. Given a convex polyhedron with exactly $n$ faces, each of which is a right triangle. Find all possible values of the integer $n$ $(n \geqslant 4)$.

Note: The angle between any two adjacent faces of a convex polyhedron is less than $180^{\circ}$.
5. If for some $n$, a polyhedron satisfying the conditions exists.

First, the total number of edges contained in each face is $3n$, and each edge is counted twice, so $2 \mid 3n$. Therefore, $n$ must be even.

For each even number $n \geqslant 4$, we construct an example that meets the conditions.

Lemma: As shown in Figure 11, let $O$ be the origin of a Cartesian coordinate system, and $X, Y$ be two different points on the plane $xOy$ (different from $O$), and $\angle OXY = 90^\circ$. Then for any point $O'$ on the $z$-axis, $\triangle O'XY$ is a right triangle, where $\angle O'XY = 90^\circ$.
Proof: Use the Pythagorean theorem to prove.
If point $O'$ coincides with $O$, no proof is needed.
If point $O'$ does not coincide with $O$, the line $OO'$ (the $z$-axis) is perpendicular to the plane $xOy$, and thus perpendicular to any line in the plane.

In the right triangles $\triangle OO'X$ and $\triangle OO'Y$, by the Pythagorean theorem, we have
$$
\begin{array}{l}
O'Y^2 = O'O^2 + OY^2 = O'O^2 + OX^2 + XY^2 \\
= O'X^2 + XY^2 \\
\Rightarrow \angle O'XY = 90^\circ.
\end{array}
$$

The lemma is proved.
If $n=4$, consider the tetrahedron $O'OXY$ in the lemma, which meets the requirements.
Below, assume $n \geqslant 6$.
Let $m = \frac{n-2}{2} \geqslant 2$.
Construct a convex $(m+2)$-sided polygon $OA_0A_1 \cdots A_m$ in the plane $xOy$ (with $O$ as the origin) such that:
(1) $OA_0 = OA_m$;
(2) All $\triangle OA_iA_{i+1} (1 \leqslant i \leqslant m-1)$ are right triangles.
As shown in Figure 12, consider $m$ rays starting from $O$ (denoted as $l_1, l_2, \cdots, l_m$, arranged clockwise). For a very small $\alpha$, we have
$$
\begin{array}{l}
\angle l_1Ol_2 = \angle l_2Ol_3 \\
= \cdots \\
= \angle l_{m-1}Ol_m = \alpha.
\end{array}
$$

Figure 12
Take any point $A_1$ on the line $l_1$. Starting from point $A_1$, draw perpendiculars from point $A_i$ to the line $l_{i+1}$, obtaining the feet of the perpendiculars $A_2, A_3, \cdots, A_m$.
Then $\angle OA_2A_1 = \angle OA_3A_2 = \cdots = \angle OA_mA_{m-1} = 90^\circ$.
From the above two equations, we know that $\triangle OA_1A_2, \triangle OA_2A_3, \cdots, \triangle OA_{m-1}A_m$ are similar.
Thus, $\frac{OA_m}{OA_{m-1}} = \cdots = \frac{OA_3}{OA_2} = \frac{OA_2}{OA_1}$.
Let this ratio be $r < 1$, where $r$ can take any value close to 1 as $\alpha$ is sufficiently small.
Notice that,
$OA_m = \frac{OA_m}{OA_{m-1}} \cdots \cdot \frac{OA_3}{OA_2} \cdot \frac{OA_2}{OA_1} \cdot OA_1 = r^m OA_1$.
Since $\alpha$ is very small, the points $A_2, A_3, \cdots, A_m$ are on the same side of the line $OA_1$.
Take point $A_0$ on the other side of $OA_1$ such that
$\angle OA_0A_1 = 90^\circ$, and $OA_0 = r^m OA_1$, where $A_0$ is the intersection of the circle with diameter $OA_1$ and the circle centered at $O$ with radius $r^m OA_1$.

If $r \rightarrow 1$ (i.e., $\alpha \rightarrow 0$), then $r^m \rightarrow 1$, ensuring that $\angle A_0OA_1$ is very small, thus the constructed polygon meets the conditions.

As shown in Figure 13, after constructing the polygon, take two points $O'$ and $O''$ on the $z$-axis (on opposite sides of the plane $xOy$) such that
$$
OO' = OO'' = OA_0 = OA_m.
$$

Consider the convex polyhedron with vertices $O', O'', A_0, A_1, \cdots, A_m$, which has exactly $n = 2m + 2$ faces, each of which is a right triangle.

In fact, by the lemma, $\triangle O'A_iA_{i+1}$ and $\triangle O''A_iA_{i+1}$ are $2m$ right triangles, and $\triangle O'A_0O''$ and $\triangle O'A_mO''$ are isosceles right triangles.