How to calculate the concentration of the elements after dissolution of iron(III) chloride?
When answering such problems, I recommend writing everything know down, then use the appropriate formula to solve for the unknown. For the first part you already did that correctly.


  In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.


\begin{align}
  m(\ce{FeCl3 (s)}) &amp;= \pu{24.7 g}\\
  M(\ce{FeCl3}) &amp;= \pu{162.3 g//mol}\\
  V(\ce{FeCl3 (aq)}) &amp;= \pu{375 mL}\\
\end{align}


  
  What is the molarity of the solution?
  


Molarity is another word for amount concentration, so you are looking for $c(\ce{FeCl3 (aq)})$.
\begin{align}
  c &amp;= \frac{n}{V}\\
  n &amp;= \frac{m}{M}\\
  c(\ce{FeCl3 (aq)}) &amp;= 
    \frac{m(\ce{FeCl3 (s)})}{M(\ce{FeCl3})\cdot V(\ce{FeCl3 (aq)})}\\
  c(\ce{FeCl3 (aq)}) &amp;= 
    \frac{\pu{24.7 g}}{\pu{162.3 g//mol}\cdot\pu{375E-3L}}\\
  c(\ce{FeCl3 (aq)}) &amp;= 
    \pu{0.406 mol//L}  
\end{align}


  
  What is the concentration of iron(III) cation in M?
  And what is the concentration of chloride anion in M? 
  


You are looking for the ion (amount) concentrations in your solution, i.e. $c(\ce{Fe^3+ (aq)})$ and $c(\ce{Cl- (aq)})$.
For these two questions you need to know how the substance dissolves. The wording here is already very explicit, as it says iron(III) cation, i.e. $\ce{Fe^3+ (aq)}$, and chloride anion, i.e. $\ce{Cl-}$. Otherwise, with some more experience you will know that this compound is a salt and it dissociates into ions when dissolved in polar solvents like water. Therefore the reaction equation is
$$\ce{FeCl3 (s) -&gt;[H2O] Fe^3+ (aq) + 3 Cl- (aq)}.$$
From this you can derive the actual relationships between the ion concentrations.
You can clearly see that there is one iron(III) ion and three chloride ions per formula of $\ce{FeCl3}$; therefore
\begin{align}
  c(\ce{Fe^3+ (aq)})&amp;= c(\ce{FeCl3 (aq)})
  &amp;&amp;=\pu{0.406 mol//L},\\
  c(\ce{Cl- (aq)}) &amp;= 3\times c(\ce{FeCl3 (aq)})
  &amp;&amp;=\pu{1.22 mol//L}.
\end{align}



The other equation would probably be appropriate when you (thermally) decompose the substance (in absence of everything) as it is the reverse equation which can be used to prepare the molecule:
$$\ce{2 Fe (s) + 3 Cl2 (g) -&gt; 2 FeCl3 (s)}.$$