I’m doing an exercise that you need to create a vector of 10 names, and you have 3 methods.
The first method is called registerName , and has the function to fill in the vectors, the second method is called sortName that returns a String (which is meant to draw names), and the third method called displayName , whose function is to display the names that have been drawn.
I have some doubts about the 3 methods, if you can help me I will thank you, I will put below parts of the code, at least the structure of it.

Here is the code with only one scope:

package Testes;

public class ClasseSorteio {

    public static void main(String[] args) {        
         String nomeSorteado[] = new String[10];




    }

    public void cadastrarNome() {
    for(int i = 0; i < 10; i++) {

    }

    }

    public String sortearNome(String nome) {
        for(int i = 0; i < 10; i++) {


        }

        return nome;
    }


    public void exibirNome() {


    }

}

In the main class I will have to use JOptionPane, and I will have to have 3 buttons: Name, Sorte Name and Show Name , each button calling a method.

I’m really sorry, if the code does not have anything at all ready, I’m really in doubt about this part of the 3 methods, so I started learning about methods now.

Thanks for the patience with this my post is not tidy.

The truth is that you did not give enough information to solve your problem, but it’s been almost a year that you’ve posted this problem, and I’ll try to give you an answer.

Arrays Indexing

It is important to mention that an array is a content store that stores multiple values. To get and set a value in the array it is necessary to refer to the position of the array where this value is / will be.

In java arrays are indexed from 0. The syntax for indexing an array is using [] . So, to get the first element of the array the statement is arr[0]

Array initialization

Java arrays can be initialized in many ways, like all other types, can be initialized in two ways:

With constant values

String nomeSorteado[] = new String[10]{
  "Pessoa1", "Pessoa2", "Pessoa3", "Pessoa4", "Pessoa5",
  "Pessoa6", "Pessoa7", "Pessoa8", "Pessoa9", "Pessoa10"
};

With dynamic values

String nomeSorteado[] = new String[10];
for(int i = 0; i < nomeSorteado.length; ++i){
   nomeSorteado[i] = <valor dinamico vai aqui>
}

These dynamic values can be obtained in several ways: User input, randomly generated values read from a data source (files, database), and so on.

Now we can solve your problem, I will resove it using constant values.

public class ClasseSorteio {
    //Nota que tive que criar um campo para poder acessar ao array em todos os metodos
    private String[] nomeSorteado = new String[10]{
      "Pessoa1", "Pessoa2", "Pessoa3", "Pessoa4", "Pessoa5",
      "Pessoa6", "Pessoa7", "Pessoa8", "Pessoa9", "Pessoa10"
    };

    public static void main(String[] args) {        
        ClasseSorteio sorteio = new ClasseSorteio();
        sorteio.cadastrarNome();
        String nome = sorteio.sortearNome();
        sorteio.exibirNome(nome);
    }

    public void cadastrarNome() {
        //não faz nada porque estou a usar valores constantes
    }

    //Nota que eu removi o parametro. 
    //Se o seu objetivo é sortear um nome no array aleatoriamente, 
    //não faz sentido passar o nome como parametro
    public String sortearNome() {
        Random r = new Random();
        int idx = r.nextInt(nomeSorteado.length);
        return nomeSorteado[idx];
    }

    //Nota que eu acrescentei o parametro.
    //Para mostrar um nome é preciso saber qual é o nome...
    public void exibirNome(String nome) {
        System.out.println(nome);
    }
}