I just need to know the net ionic equation and balance them. I figured the equation. but the net and balance is harder.

Fe(NO3)3 (aq) + Na2CO3 (aq) = FeCO3 (s) + Na2(NO3)3 (aq)

Fe(NO3)3 (aq) + NaI (aq) =FeI (s) + Na(NO3)3 (aq)

Ba(NO3)3 (aq) + Na3PO4(aq) = BaPO4 (s) + Na3(NO3)3 (aq)

Here is how you do them.

solids are written as the molecule.
aq are written as the ions.
gases as molecules
weak electrolytes (weak acids, weak bases, water) as molecules.
Then cancel ions common to both sides.

However, before you begin that you must learn how to write and balance the equation. I will do the first one for you. REMEMBER THIS. After you write the formula you may NOT change the subscripts I can tell that you are trying to balance the equation at the time you write the equation. These are two separate steps.
You may be transcribing a word equation to a chemical equation. Write those as a separate post and we can try to get things right at the beginning.

To balance these equations and determine the net ionic equation, you will need to follow a few steps:

1. Write the skeleton equation: In this step, you have already written the skeleton equations as provided:

Fe(NO3)3 (aq) + Na2CO3 (aq) = FeCO3 (s) + Na2(NO3)3 (aq)
Fe(NO3)3 (aq) + NaI (aq) = FeI (s) + Na(NO3)3 (aq)
Ba(NO3)3 (aq) + Na3PO4(aq) = BaPO4 (s) + Na3(NO3)3 (aq)

2. Separate the soluble (aqueous) compounds: Identify the compounds that are aqueous (dissolved in water) and those that are solid (precipitate).

In the first equation:
Fe(NO3)3 (aq) + Na2CO3 (aq) = FeCO3 (s) + Na2(NO3)3 (aq)

Fe(NO3)3 and Na2(NO3)3 are both soluble and remain in the solution.
FeCO3 is insoluble and forms a solid precipitate.

In the second equation:
Fe(NO3)3 (aq) + NaI (aq) = FeI (s) + Na(NO3)3 (aq)

Fe(NO3)3 and Na(NO3)3 are soluble and remain in the solution.
FeI is insoluble and forms a solid precipitate.

In the third equation:
Ba(NO3)3 (aq) + Na3PO4(aq) = BaPO4 (s) + Na3(NO3)3 (aq)

Ba(NO3)3 and Na3(NO3)3 are soluble and remain in the solution.
BaPO4 is insoluble and forms a solid precipitate.

3. Write the balanced complete ionic equation: This equation shows all the ions that are present in the reaction.

In the first equation:
Fe^3+(aq) + 3NO3^-(aq) + 2Na+(aq) + CO3^2-(aq) → FeCO3 (s) + 3Na+(aq) + 3NO3^-(aq)

In the second equation:
Fe^3+(aq) + 3NO3^-(aq) + Na+(aq) + I^-(aq) → FeI (s) + Na+(aq) + 3NO3^-(aq)

In the third equation:
Ba^2+(aq) + 3NO3^-(aq) + 3Na+(aq) + PO4^3-(aq) → BaPO4 (s) + 3Na+(aq) + 3NO3^-(aq)

4. Determine the net ionic equation: The net ionic equation shows only the essential reaction components, excluding spectator ions that remain unchanged. Cancel out the identical ions that appear on both sides of the equation.

In the first equation:
Fe^3+(aq) + CO3^2-(aq) → FeCO3 (s)

In the second equation:
Fe^3+(aq) + I^-(aq) → FeI (s)

In the third equation:
Ba^2+(aq) + PO4^3-(aq) → BaPO4 (s)

These net ionic equations represent the essential reactions occurring in each equation, omitting the spectator ions.