Moles of chlorine gas needed to make up an equivalent amount of saturated salt solution
a) molar concentration
There are indeed 6.14 moles of $\ce{NaCl}$ in 359 grams of $\ce{NaCl}$.
I'd prefer to write the molar mass as &quot;$\pu{58.44 g mol^-1 }\ \ce{NaCl}$&quot; or &quot;$\pu{58.44 g/mol}\ \ce{NaCl}$&quot; to keep it all together.
b) molal concentration
You got the right value, but you did it the wrong way. For the solvent:
$$\dfrac{\pu{980 kg/m^3}}{\pu{1000 L/m^3}} = \pu{0.980 kg/L}$$
c) moles of $\ce{Cl2}$ gas
The book is correct. There are 6.14 moles of $\ce{NaCl}$. Each molecule of chlorine gas contains two atoms of chlorine, so you need half as many moles to get the same number of atoms of Cl. So 6.14/2 = 3.07
This part of the question has poor phrasing. You can't dissolve 3.07 moles of $\ce{Cl2}$ in water and get a solution of $\ce{NaCl}$.
d) density of the mixture
Your answer is correct.