= Solution
Normalize the supplied <Bell state> as $|\Psi^+\rangle=(|01\rangle+|10\rangle)/\sqrt2$, with $|V\rangle=|0\rangle$ and $|H\rangle=|1\rangle$. For a <measurement of linear photon polarization>, the two orthogonal analyzer states can be taken as
$$
|\phi,+\rangle=\cos\phi|0\rangle+\sin\phi|1\rangle,\qquad
|\phi,-\rangle=-\sin\phi|0\rangle+\cos\phi|1\rangle.
$$
Assigning outcomes $+1,-1$ gives the <observable>
$$
M(\phi)=|\phi,+\rangle\langle\phi,+|-|\phi,-\rangle\langle\phi,-|
=\begin{pmatrix}\cos2\phi&\sin2\phi\\\sin2\phi&-\cos2\phi\end{pmatrix}
=\cos2\phi\,Z+\sin2\phi\,X,
$$
where $X,Z$ are the <Pauli operators>. The angle of the <Bloch vector> is twice the physical analyzer angle. Hence
$$
\boxed{A_1=Z,\quad A_2=X,\quad A_3=\frac{Z+X}{\sqrt2};\qquad
B_1=Z,\quad B_2=\frac{Z-X}{\sqrt2},\quad B_3=\frac{Z+X}{\sqrt2}.}
$$
Each has <eigenvalues> $\pm1$ and square equal to the <identity operator>.
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