Unitary matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are on the unit circle. Hermitian matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are real. So unitary Hermitian matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are $\pm 1$.
This is a very strong condition. As George Lowther says, any such matrix $M$ has the property that $P = \frac{M+1}{2}$ admits a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are $0, 1$; thus $P$ is a Hermitian idempotent, or as George Lowther says an orthogonal projection. Of course such matrices are interesting and appear naturally in mathematics, but it seems to me that in general it's more natural to start from the idempotence condition.
I suppose one could say that Hermitian unitary matrices precisely describe unitary representations of the cyclic group $C_2$, but from this perspective the fact that such matrices happen to be Hermitian is an accident coming from the fact that $2$ is too small.
First you have to find an orthonormal basis of eigenvectors of $A$. Just find any eigevectors, they are necessarily orthogonal because you have three eigenspaces, and distinct eigenspaces are always orthogonal to each other. Then normalize your vectors to get an orthonormal basis $v_1,v_2,v_3$ for eigenvalues $0,1,-1$. Let $V=(v_1|v_2|v_3)$ be the orthogonal matrix with columns $v_1,v_2,v_3$. Then $D:=V^*AV$ is the diagonal matrix with diagonal values $0,1,-1$. Do the same for $B$ to get $u_1,u_2,u_3$, and $D=U^*BU$. Then $A=VU^*BUV^*=(UV^*)^*B(UV^*)$ with orthogonal and therefore unitary $UV^*$.
We have $v_1,v_2,v_3=(0,1,0),(1,0,1)/\sqrt{2}, (1,0,-1)/\sqrt{2}$ and $u_1,u_2,u_3=(1,-1,0)/\sqrt{2}, (1,1,0)/\sqrt{2}, (0,0,1)$.
Best Answer
$|\det(A+B)| = |\det(A(I+A^{-1}B))| = |\det(I+A^{-1}B)|$. Now, $A^{-1}B$ is unitary, so it's eigenvalues lie on the unit circle $B_1(0)$. Thus the eigenvalues of $I+A^{-1}B$ lie in $1+B_1(0)\subset B_2(0)$. I think, that's all you need.