Density of a sodium hydroxide solution
What's your first miss is the definition of $50\%(w/w)$ $\ce{NaOH}$ solution (although it doesn't matter here). Actual definition is $\pu{50 g}$ of $\ce{NaOH}$ in $\pu{100 g}$ of solution. Since water is the solvent, your interpretation is correct.
Your second mistake is the lethal one. You have considered water and solid $\ce{NaOH}$ are additive. Assuming $V_\text{total} = V_\ce{NaOH} + V_\ce{H2O}$ is a mistake.
Using a given density of $\ce{NaOH}$ ($\pu{1.5298 g}$),we can calculate the volume of $\pu{1000 g}$ of solution: $\frac{\pu{1000 g}}{\pu{1.5298 g/mL}} = \pu{653.68 mL}$. Therefore, dissolving $\pu{500 g}$ of $\ce{NaOH}$ in $\pu{500 mL}$ of water (assuming temperature is $\pu{25 ^\circ C}$, hence density of water is $\pu{1.00 g/mL}$), the volume increased by only $\pu{153.68 mL}$. Do you still think volumes are additive?