Predicting the products of the radical chlorination of propane
Let me give you two helpful hints.


  Propane


If you use an internet search tool of your choice, you should quickly find out that propane is $\ce{C3H8}$, which has the structure $\ce{CH3CH2CH3}$. The formula and structure you are using is octane. The product you drew, 1-chlorooctane $\ce{CH3CH2CH2CH2CH2CH2CH2CH2Cl}$, is one of the monochlorination products of octane, so you should feel good about being able to correctly apply the reaction type and produce a correct product.


  What is the question asking?


A careful read of the intro part of the question When propane reacts with chlorine, two different monochloropropanes are formed indicates that we should be expecting two different organic products with chlorine atoms attached to the carbon skeleton. In other words this reaction should give two products with the formula $\ce{C3H7Cl}$. 

However, part a) is not worded well to indicate that you mostly likely are to draw the structures of the two monochloropropane products. You are correct in noting that a radical halogenation reaction will have two products, though the inorganic byproduct is usually considered to be $\ce{HCl}$ and not $\ce{H2}$ or $\ce{H2O}$. 

$$\ce{C3H8 + Cl2 -&gt; C3H7Cl + HCl}$$

So let me ask your question in a clearer way: 


  Propane reacts with chlorine to produce two different monochlorination products with the formula $\ce{C3H7Cl}$. (a) Draw the structures of  both monochlorination products and provide their names. (b) Identify the type of isomeric relationship that exists between the two structures that you drew in (a).