= Solution
For $\mathbf b\ne0$, define the <orthogonal projection> of $\mathbf a$ onto its span by
$$
\mathbf a_{\parallel}=\frac{\mathbf a\cdot\mathbf b}{|\mathbf b|^2}\mathbf b,\qquad \mathbf a_{\perp}=\mathbf a-\mathbf a_{\parallel}.
$$
The first is parallel or antiparallel to $\mathbf b$, or zero, and $\mathbf a_\perp\cdot\mathbf b=0$. This proves the decomposition. If $\mathbf a\ne0$ as well, comparison with $\mathbf a_\parallel=|\mathbf a|\cos\theta\,\widehat{\mathbf b}$ yields
$$
\boxed{\cos\theta=\frac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}.}
$$
Indeed $|\mathbf a|^2=|\mathbf a_\parallel|^2+|\mathbf a_\perp|^2$ shows that the ratio lies in $[-1,1]$. There is a unique $\theta\in[0,\pi]$ with this cosine. The definition is symmetric in the two <vectors>, unchanged by orthogonal coordinate changes and positive rescaling, and agrees with ordinary plane geometry in their span: projecting a unit vector onto the other gives its cosine. Parallel, perpendicular and antiparallel directions give $0,\pi/2,\pi$ respectively. The zero vector has no defined angle; if $\mathbf b=0$, the nonzero-axis projection formula itself must not be used.
The vertices of the centered <hypercube> are sign vectors $\mathbf u\in\{-1,1\}^n$. A body diagonal identifies $\mathbf u$ and $-\mathbf u$, hence there are $2^{n-1}$ such diagonals. If two sign vectors disagree in $h$ positions, their <dot product> is $n-h-h=n-2h$. For odd $n$ this integer is odd and cannot be zero. Thus \b[no two body diagonals are perpendicular in odd dimension], by the <orthogonality of hypercube body diagonals>.
In four dimensions, mutually <orthogonal> nonzero <vectors> are linearly independent, so at most four mutually perpendicular diagonals are possible. Four are achieved by the directions
$$
\boxed{(1,1,1,1),\quad(1,-1,1,-1),\quad(1,1,-1,-1),\quad(1,-1,-1,1).}
$$
Each has squared norm four and every distinct pair has zero <dot product>; these are the rows of a <Hadamard matrix>. Therefore \b[the maximum is four].
For distinct diagonals, the <Hamming distance> is $h=1,2,3$; $h=0,4$ would represent the same line. Oriented vertex directions therefore give cosines $1/2,0,-1/2$. The angles are
$$
\boxed{\pi/3,\ \pi/2,\ 2\pi/3\quad\text{for oriented directions}.}
$$
For unoriented diagonals, the usual smaller angle between the lines is \b[$\pi/3$ or $\pi/2$]; the supplementary intersection angle $2\pi/3$ also occurs when the two rays are chosen oppositely. Stating this convention avoids counting a line and its reverse as different diagonals.
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