= Solution
The scalar potential is
$$
V=-\frac{\mu^2}{2}(\phi_1^2+\phi_2^2)
+\frac\lambda4(\phi_1^2+\phi_2^2)^2
=\frac\lambda4\left(\phi_1^2+\phi_2^2-\frac{\mu^2}{\lambda}\right)^2
-\frac{\mu^4}{4\lambda}.
$$
Since $\mu^2,\lambda>0$, all minima form the <circle> $\phi_1^2+\phi_2^2=v^2$, with $v=\mu/\sqrt\lambda>0$. Choose the representative <vacuum> $\langle\phi_1\rangle=v$, $\langle\phi_2\rangle=0$, and write $\phi_1=v+h$, $\phi_2=\pi$. A nontrivial scalar rotation moves this <vacuum> to another minimum, so the continuous symmetry undergoes <spontaneous symmetry breaking>. The invariant action and the selected ground state have different stabilizers; the <circle> of equivalent minima is not evidence that the action lost its symmetry.
The Yukawa terms become
$$
\mathcal L_Y=-gv\bar\psi\psi-gh\bar\psi\psi-ig\pi\bar\psi\gamma_5\psi.
$$
Comparison with the free <Dirac equation> gives
$$
\boxed{m_\psi=gv\ \text{as the signed mass coefficient},\qquad
m_{\rm physical}=|g|v.}
$$
For the usual choice $g>0$, this is simply $gv$. The absolute value matters if the real coupling is negative: the dispersion depends on $m_\psi^2$, and a field redefinition can change the <mass> coefficient's sign. There is no explicit symmetry-breaking <fermion> <mass> parameter in the original theory.
The scalar symmetry is broken completely. Because the <spinor> transformation uses a half angle, $\alpha=2\pi$ leaves the scalars fixed while sending $\psi\to-\psi$; this discrete <fermion> parity is unbroken in the full combined transformation group. It does not remove the broken continuous generator or the associated <Goldstone boson>.
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