= Solution
Write $m_i=\mu_i^2$ and $r^2=\phi_1^2+\phi_2^2$. The uniform <Landau free energy> is
$$
V(\phi_1,\phi_2)=\frac12m_1\phi_1^2+\frac12m_2\phi_2^2+gr^4,
$$
so stationarity requires
$$
\phi_i(m_i+4gr^2)=0.
$$
For $m_1,m_2>0$, the unique <ground state> is $(0,0)$ and the $\mathbb Z_2\times\mathbb Z_2$ <discrete symmetry> is unbroken. If $m_1<0$ and $|m_1|>|m_2|$, then
$$
\boxed{(\phi_1,\phi_2)=\left(\pm\sqrt{\frac{-m_1}{4g}},0\right),}
$$
which breaks the first $\mathbb Z_2$ and leaves the second intact. This includes the case in which both masses are negative, because $m_1$ is then the more negative one. The remaining possible ordered case under the stated inequality is $m_1>0>m_2$, for which
$$
\boxed{(\phi_1,\phi_2)=\left(0,\pm\sqrt{\frac{-m_2}{4g}}\right),}
$$
and only the second $\mathbb Z_2$ is broken. The <Hessian matrix> in each ordered state is positive because the uncondensed direction has squared mass $m_k-m_j>0$, where $m_j$ is the condensed, more negative mass.
Near either continuous transition the nonzero order parameter is proportional to $(-m_i)^{1/2}$. Hence the <mean-field critical exponents> are
$$
\boxed{\beta_1=\beta_2=\frac12.}
$$
The <lower critical dimension> is the dimension at or below which fluctuations destroy the proposed finite-temperature ordered phase. Here the broken symmetry is discrete, so $d_{\rm l}=1$. In one dimension a <domain wall> interpolating between the two signs has finite energy, whereas its possible position gives an <entropy> growing as $\log L$. Domain walls therefore occur with nonzero density at every positive <temperature> and split the system into domains of finite typical length. Thus \b[there is no finite-temperature ordered phase in one dimension].
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