Assign $u$ to be the unit normal vector to some plane $S$ in $\mathbb{R}^3$, then $u\times x_n \in T(S)$ for any $n$, which means $u\times x_n $ is a vector field on the plane $S$. After the first cross product, all $x_n$'s ($n > 1$) lie on the plane $S$, because $x_{n}\cdot u = (u\times x_{n-1})\cdot u = 0$ for all $n> 1$.
Now $x_{n+2} = u\times(u\times x_n)$, for $n>1$ where $x_n$ already is a vector field on this plane, geometrically speaking, doing cross product twice is the same as rotating $x_n$ twice counter-clockwisely with respect $u$ by $\pi/2$, and you get $-x_n$.
Proof-wise, use the cross product definition $u\times x_n = |u||x_n|\sin(\theta) \boldsymbol{\nu}_n$, where $\theta$ is the angle between $u$ and $x_n$, for $n>1$, this is $\pi/2$, because $x_{n}\cdot u =0$ for all $n> 1$ which is proved above, and dot product of two vectors is the cosine of the angle between them if normalized by their norms.
$\boldsymbol{\nu}_n$ is the unit vector perpendicular to the plane containing $u$ and $x_n$ directioning by right-hand rule, recalling that $x_n$ is already a vector field on the plane $S$ that has unit normal $u$, $\boldsymbol{\nu}_n$ is in the direction of the counter-clockwise $\pi/2$-rotation of $x_n$ on this plane. Hence $u\times x_n = |x_n| \boldsymbol{\nu}_n$.
Next we have $u\times(u\times x_n) = |x_n| u\times\boldsymbol{\nu}_n = |x_n|\,|u|\, |\boldsymbol{\nu}_n|\,\sin(\theta)\boldsymbol{\nu}_{n+1}$, $\theta$ is again $\pi/2$ due to the same reason that the dot product is zero, and $\boldsymbol{\nu}_{n+1}$ is now the unit vector perpendicular to the plane containing $u$ and $u\times x_n$ directioning by right-hand rule, repeat the right hand rule argument above, we know that $\boldsymbol{\nu}_{n+1}$ is in the direction of $x_n$'s counter-clockwise $\pi$-rotation with respect to $u$, therefore $u\times(u\times x_n) = |x_n|\boldsymbol{\nu}_{n+1} = -x_n$.
Remark: In vector calculus, the projection of any pointwisely well-defined vector field $\boldsymbol{v}$ onto a smooth surface $S$ is defined as $\boldsymbol{n}\times(\boldsymbol{v}\times\boldsymbol{n})$, where $\boldsymbol{n}$ is the outer unit normal of this surface, now if $x_n$ is already a vector field on $S$, its projection is itself.
For the second statement, apply the first $x_5 = -x_3 = u\times(x_1\times u)$, either repeat the argument above, or directly use above remark.
Hence both statements are true, given that $u,x_1 \in \mathbb{R}^3$.