= Solution
First $E=\ker(A+I)$ is a <linear subspace>. To prove it is nonzero, use <orthogonality> and $\det A=-1$:
$$
\det(A+I)=\det A\,\det(I+A^{-1})
=-\det(I+A^T)=-\det(A+I).
$$
Hence $\det(A+I)=0$. If $u\in E$ then $Au=-u$ also implies $A^Tu=-u$. For $x\perp E$,
$$
\langle Ax,u\rangle=\langle x,A^Tu\rangle=-\langle x,u\rangle=0,
$$
so $E^\perp$ is invariant. If $E$ had dimension two, its one-dimensional <orthogonal complement> would be invariant and $A$ would act there as a scalar $\eta=\pm1$. The <determinant> would be $(-1)^2\eta=-1$, so $\eta=-1$, forcing $A=-I$. Dimension three also forces $A=-I$. Both contradict the hypothesis. Consequently
$$
\boxed{\dim E=1.}
$$
Choose an <orthonormal basis> beginning with a <unit vector> in $E$ and followed by a basis of $\Pi=E^\perp$. In this basis the <orthogonal matrix> has block form
$$
A\sim\begin{pmatrix}-1&0\\0&R\end{pmatrix},
\qquad R^TR=I_2,\qquad \det R=1.
$$
The first column of $R$ is some $(\cos\theta,\sin\theta)^T$. Its perpendicular unit second column with positive orientation must be $(-\sin\theta,\cos\theta)^T$. Thus the restriction is the plane <rotation>
$$
R=\begin{pmatrix}\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\end{pmatrix}.
$$
<Trace> invariance gives $\operatorname{tr}A=-1+2\cos\theta$, proving
$$
\boxed{\cos\theta=\tfrac12(\operatorname{tr}A+1).}
$$
A <Euclidean reflection> in $\Pi$ fixes that plane pointwise and reverses $E$, so it occurs exactly when $R=I_2$. This is equivalent to $\cos\theta=1$, hence \b[exactly to $\operatorname{tr}A=1$]. Finally,
$$
\det(A-I)=(-2)\det(R-I_2)
=-2\bigl((\cos\theta-1)^2+\sin^2\theta\bigr)
=\boxed{4(\cos\theta-1)}.
$$
This exhibits the <three-dimensional improper orthogonal transformation> as a <rotation> about the axis $E$ composed with <Euclidean reflection> in $\Pi$.
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