Fred Kleinschmidt

Presumably sc is a Scanner. So sc.nextLine() returns a String from that scanner.
"Double" is a Java class; it has a static method called parseDouble.
Static methods are called using the form "ClassName.methodName()".
So Double,parseDouble( Strikng string ) is invoking the Double class static method which takes a String as its argument and returns a double value if the string can be interpreted as representing a double value, or it throws an exception if the string does not represent a double value.

Which of the alternatives is correct?

Which one do you think is correct, and why? Why do you think the others are incorrect?

One way without switch is to not include the same test condition when unnecessary:
if( ageIsFiltered ) {   // add all tests for ageFiltered   if  (genderIsFiltered) {      //add code here   } else {      // add code   }   ... } else {   // Add tests for not ageFiltered   ... }

The program repeats until the input string is quit 0.

This means, to me, that it should exit only if the input is "quit 0"
       while (  (num != 0) || !word.equals("quit" ) )        {            System.out.printf("Eating %d %s a day keeps the doctor away.\n",num,word);            word = user.next();                      num = user.nextInt();        }

In addition to what Stephan wrote, do you know what a tree is? What is the difference between a node and a leaf?
And the given solution assumes that the zeroth entry in the array is an object of type Integer. It should not be cast to Integer without first assuring that such a cast is legal.

An Object[] is a very poor choice to implement a tree.

Actually, if you are only using %s as the format (and not something more complicated, for example for real  number formatting) I would just use:
System.out.println( "Name: " + imena[rand.nextInt(imena.length)] ); System.out.println( "Surname: " + prezimena[rand.nextInt(prezimena.length)] ); System.out.println( "Position: " +  pozicije[rand.nextInt(pozicije.length)]);
It is easier to read, and easier to modify if needed.

Get rid of the second % in each specifier., ...
for example, "Name: %s\nSurname: %s\nPosition: %s\n", ...

For selecting based on probability, generate a random integer between 1 and 100 inclusive.
Then if the value is in the range 1 to 10 (10 %), select "Bad".
If it is between 11 and 70 (60 %), pick "normal".
If it is between 71 and 90 (20 %), pick "good".
Otherwise (i.e., from 91 to 100) (10 %), choose "Excellent".

You do not need to use pow() to solve this. Each square contains twice the previous value.

The first square contains f(1) =2^0=1, the second contains f(2) = 2^1=2 = 2*f(1), the nth contains f(n) = 2^(n-1) = 2 * f(n-1)  so the 64th square contains 2^63.

As Campbell Ritchie pointed out, this is larger than will fit in an int or long (The maximum ling is 2^63 - 1).

You haven't said what your difficulty is. Is it  not working? If not, what is it doing instead of what yu want?

Note that you should never delay or sleep inside a method that is executed on the EventDispatchTread.

Using the terms "Parent class" and "Child class" is a poor choice of terminology. One should use "Superclass" and "Subclass" instead.

"Parent" and "Child" are usually used  to describe GUI relationships.
For example, when you add a JLabel to a JPanel, the panel is the parent of the label, and the label is a child of the panel (see java.awt.Component.getParent()).
So the label is a child of the panel, but it is not a subclass of JPanel.

Similarly, if you create a new class MyLabel that is a subclass of JLabel, and add a myLabel to the JPanel, the myLabel's parent is the JPanel. JLabel is the superclass of MyLabel, not its parent.

Note that
27%26 = 1
and
26%26 = 0
So you will need to handle the case where the answer is zero differently if you want the answer to be 26.

Also, the do-loop ending at line 83 with this criterion: while( 1 > 0 );seems to be an infinite loop with no break statement

In general, it is easier to solve if you rotate both boxed so that they lie flat in the x-y plane, where it is easier to calculate the volume of each. Then see if the one with the smaller volume will fit in the larger box.

if((month == 1 || month == 3 || month == 5 || month == 7 ||
          month == 8 ||month == 10 || month == 12 )||day<=1||day>=31){
          _day=day;}

If the user enters 0 for day, this line will set _day to 0 regardless of the month.

Perhaps you mean to use something like
if ( month has 31 days ) {   if (  (day > 0) && (day <= 31) ) {      _day = day;   } else {      _day = 1;   } }

You never compare second and third (your third if-statement is incorrect).

Here's one way to do it:
int answer = ( first >= second ) ? first : second; if ( answer < third ) {   return third; } return answer;