Here is a quick sketch of the solutions.
In grading, I am looking only for a few things on each question,
and an overall understanding of the important concepts.

1. (a) O( N ) because of the single loop, and the calls to get are O(1)
   (b) O( N^2 ) because of the single loop, and each get is O(N)time
   (c) For a linear algorithm, when N is increased
       by a factor of 5, the running time increases
       by a factor of 5. So the time will be 20 sec.
   (d) Use an iterator to get an O( N ) algorithm for all types of lists.

2. (a) All operations are O(N) because getNode is O(N) and the
       other two invoke getNode.

   (b)
     private Node<AnyType> getNode( int idx )
     {
         if( idx < 0 )
             throw new IndexOutOfBoundsException( );

         Node<AnyType> ptr = header.next;

         for( int i = 0; i != idx; i++, ptr = ptr.next )
             if( ptr == null )
                 throw new IndexOutOfBoundsException( );

         return ptr;
    }

   (c)
     public void remove( int idx )
     {
         Node<AnyType> p = getNode( idx );

         if( p == tail )
             throw new IndexOutOfBoundsException( );

         p.prev.next = p.next;
         p.next.prev = p.prev;
     }

   (d)
     public void add( int idx, AnyType x )
     {
         Node<AnyType> p = getNode( idx );
         p.prev = p.prev.next = new Node<AnyType>( x, p.prev, p );
     }

3.

   (a)
     public static List<String> duplicateCodes( String [ ] arr )
     {
         Map<String,Integer> codes = new HashMap<String,Integer>( );

         for( String id : arr )
         {
              String thisCode = "XXX" + id.substring( 3 );

              Integer count = codes.get( thisCode );
              if( count == null )
                  codes.put( thisCode, 1 );
              else
                  codes.put( thisCode, count + 1 );
         }

         List<String> result = new ArrayList<String>( );
         for( Map.Entry<String,Integer> e : codes.entrySet( ) )
             if( e.getValue( ) >= 2 )
                 result.add( e.getKey( ) );

         return result;
     }
             
              

4.
     public static File findMax( File d )
     {
         File maxFile = d;

         if( d.isDirectory( ) )
             for( File f : d.listFiles( ) )
             {
                 File thisFile  = findMax( f );
                 if( thisFile.length( ) > maxFile.length( ) )
                     maxFile = thisFile;
             }

         return maxFile;
     }