Balance the oxidation reduction equation in half reaction

PbO2 + KCl => KClO + KPb(OH)3
thank you a lot

You're right. I turned it around. Sorry about. Just reverse everything and you will have that half reacted.

For KPb(OH)3 the cation is K^+ and the anion is Pb(OH)3^-
So I would write
PbO2 ==> Pb(OH)3^-
I assume this is a basic solution.
Pb is +4 on the left and +2 on the right.
PbO2 + 2e + 2H2O ==> Pb(OH)3^- + OH^-
You may add K^+ to each side to make it read
PbO2 + 2e + 2H2O + K^+ ==> KPb(OH)3 + OH^-
but I wouldn't add K^+ until both half reactions have been balanced, multiplied by the appropriate number and added together. Then add positive or negative ions as needed to make compounds and not ions.
BTW, I found your follow up question about the 2.48 g salt, endothermic and exothermic and responded to that post. If you have trouble finding it I can find it and post it where you like. And while I'm on the subject, that's another very good reason for using the same screen name. Had I known that student was the same as this one I could have taken care of that earlier.

In acid solution.?

2H^+ + ClO^- + 2e ==> Cl^- + H2O

In basic solution?
H2O + ClO^- + 2e --> Cl^- + 2OH^-

The way to learn this is to do them. Looking at what someone else has done won't get it. If you will share what your problem(s) is/are in doing these perhaps I can help you over your "trouble" spots.

But ?

but in the main reaction , the ClO^- is on the product site not reactant , as you put it on the reactant site , I mean Cl- is oxidized to ClO^- !

On the other hand , I can balance it by myself , but I'm having trouble with KPb(OH)3 , I don't know how to separate it into its cation and anion

To balance the oxidation-reduction equation, we need to make sure that the number of atoms and charges are equal on both sides of the equation. Here's how you can balance this half-reaction:

1. Separate the oxidation and reduction half-reactions:
Oxidation: PbO2 -> KClO
Reduction: KCl -> KPb(OH)3

2. Balance the atoms, starting with those that appear in more than one species:
Oxidation: PbO2 -> KClO
Pb: 1 Pb on the left, 1 Pb on the right
O: 2 O on the left, 1 O on the right
Cl: 0 Cl on the left, 1 Cl on the right (add KCl as a spectator)

Reduction: KCl -> KPb(OH)3
K: 1 K on the left, 1 K on the right
Cl: 1 Cl on the left, 0 Cl on the right (add KCl as a spectator)
H: 0 H on the left, 3 H on the right (add H2O as needed)
O: 0 O on the left, 3 O on the right (add H2O as needed)

3. Balance the charges by adding electrons (e⁻):
Oxidation: PbO2 -> KClO + 2e⁻
Reduction: 2Cl⁻ -> KPb(OH)3 + 6e⁻

4. Multiply each half-reaction by integers to equalize the number of electrons:
Oxidation: 3PbO2 -> 3KClO + 6e⁻
Reduction: 6Cl⁻ -> 3KPb(OH)3 + 18e⁻

5. Combine the two half-reactions and cancel out any spectator ions:
3PbO2 + 6KCl + 6H2O -> 3KClO + 3KPb(OH)3

Finally, the balanced oxidation-reduction equation is:
3PbO2 + 6KCl + 6H2O -> 3KClO + 3KPb(OH)3.