Disproportionation Reactions of the Halogens: Writing balanced net ionic equations for four reactions. See details I first conducted the following reactions, each in separate test tubes:

$\ce{NaBr}$ + $\ce{H2SO4}$ + $\ce{NaBrO3}$
$\ce{KI}$ + $\ce{H2SO4}$ + $\ce{KIO3}$

...with water and dichloroethane added to each as well.
(In the experiment, I put a small quantity of sodium bromide in one test tube and a similar quantity of potassium iodide in the other test tube. I then added 1 mL of water, 3 drops of 9M sulfuric acid, and 2 mL of dichloroethane to each. I then added a small quantity of sodium bromate to the first test tube and a similar quantity of potassium iodate to the other.)
Then, I removed the aqueous layer from each test tube and added $\ce{NaOH}$ to the dichloroethane layer in each.

(#1) $\ce{NaOH}$ added to dichloroethane
(#2) $\ce{NaOH}$ added to dichloroethane

So far, I have the following equations written:

$\ce{NaBrO3 + 5NaBr + 3H2SO4 -&gt; 3Na2SO4 + 3H2O + 3Br2}$
$\ce{KIO3 + 5KI + 3H2SO4 -&gt; 3K2SO4 + 3H2O + 3I2}$
$\ce{3Br2 + 6OH^{-} -&gt; 5Br^{-} + BrO3^{-} + 3H2O}$
$\ce{3I2 + 6OH -&gt; 5I^{-} + IO3^{-} + 3H2O}$

Please check my work. If without error, I just need to convert these to equations to net ionic.
(answer to original question)

No, you are not on the right track. 

If bromate is a starting material, it should not appear as a product in reaction 1. Instead, consider how bromide and bromate can react with each other.

If iodate is a starting material, it should not appear as a product in reaction 2. Instead, consider how iodide and iodate can react with each other.

For reactions 3 and 4, a hint would be "disproportionation".