What does m± represent in the equation for mean activity of a salt solution?
The top $m$ is the mean ionic molality of the solution you have.  You must know the concentration since you need it to compute the activity coefficient.  The bottom $m_0$ is the standard molality which in general is $\pu{1 M}$.  It is there to make the activity dimensionless.

The required formula for $m_\pm$ is $$m_\pm^\nu = (m_+)^{\nu_+} (m_-)^{\nu_-}$$ where $m_-$ is the molality of the negative ions and $m_+$ is the molality of the positive ions. Also $\nu_-$ is the stoichiometric coefficient of the negative ions and $\nu_+$ is the same for the positive ions. Also $\nu = \nu_- + \nu_+$.

Thus in your specific case, where you have $\pu{0.0120 M}$ of $\ce{Na3PO4}$, we have

\begin{align}
\nu_+ &amp;= 3 &amp; \nu_- &amp;= 1 &amp; \nu &amp;= 4 \\
m_+ &amp;= \pu{0.0360 M} &amp; m_- &amp;= \pu{0.0120 M}
\end{align}

and hence

\begin{align}
m_\pm &amp;= \left[(\pu{0.0360 M})^3(\pu{0.0120 M})^1\right]^{1/4} \\
&amp;= \pu{0.0274 M}
\end{align}