Digital Outputs along with Shift Registers?

I am using a 595 shift reg to control 8 LED’s and have 2 more LED’s to control.

Using the example code:

/*
  Shift Register Example
  Turning on the outputs of a 74HC595 using an array

 Hardware:
 * 74HC595 shift register 
 * LEDs attached to each of the outputs of the shift register

 */
//Pin connected to ST_CP of 74HC595
int latchPin = 8;
//Pin connected to SH_CP of 74HC595
int clockPin = 12;
////Pin connected to DS of 74HC595
int dataPin = 11;

//holders for infromation you're going to pass to shifting function
byte data;
byte dataArray[10];

void setup() {
  //set pins to output because they are addressed in the main loop
  pinMode(latchPin, OUTPUT);
  Serial.begin(9600);

  //Binary notation as comment
  dataArray[0] = 0xFF; //0b11111111
  dataArray[1] = 0xFE; //0b11111110
  dataArray[2] = 0xFC; //0b11111100
  dataArray[3] = 0xF8; //0b11111000
  dataArray[4] = 0xF0; //0b11110000
  dataArray[5] = 0xE0; //0b11100000
  dataArray[6] = 0xC0; //0b11000000
  dataArray[7] = 0x80; //0b10000000
  dataArray[8] = 0x00; //0b00000000

  //function that blinks all the LEDs
  //gets passed the number of blinks and the pause time
  blinkAll_2Bytes(2,500); 
}

void loop() {

  for (int j = 0; j <= 8; j++) {
    //load the light sequence you want from array
    data = dataArray[j];
    //ground latchPin and hold low for as long as you are transmitting
    digitalWrite(latchPin, 0);
    //move 'em out
    shiftOut(dataPin, clockPin, data);
    //return the latch pin high to signal chip that it 
    //no longer needs to listen for information
    digitalWrite(latchPin, 1);
    Serial.println("No. 1");
    Serial.println(j);
    delay(300);
  }
  for (int i = 8; i >= 0; i--) {
    //load the light sequence you want from array
    data = dataArray[i];
    //ground latchPin and hold low for as long as you are transmitting
    digitalWrite(latchPin, 0);
    //move 'em out
    shiftOut(dataPin, clockPin, data);
    //return the latch pin high to signal chip that it 
    //no longer needs to listen for information
    digitalWrite(latchPin, 1);
    Serial.println("No. 2");
    Serial.println(i);
    delay(300);
  }
}



// the heart of the program
void shiftOut(int myDataPin, int myClockPin, byte myDataOut) {
  // This shifts 8 bits out MSB first, 
  //on the rising edge of the clock,
  //clock idles low

  //internal function setup
  int i=0;
  int pinState;
  pinMode(myClockPin, OUTPUT);
  pinMode(myDataPin, OUTPUT);

  //clear everything out just in case to
  //prepare shift register for bit shifting
  digitalWrite(myDataPin, 0);
  digitalWrite(myClockPin, 0);

  //for each bit in the byte myDataOut�
  //NOTICE THAT WE ARE COUNTING DOWN in our for loop
  //This means that %00000001 or "1" will go through such
  //that it will be pin Q0 that lights. 
  for (i=7; i>=0; i--)  {
    digitalWrite(myClockPin, 0);

    //if the value passed to myDataOut and a bitmask result 
    // true then... so if we are at i=6 and our value is
    // %11010100 it would the code compares it to %01000000 
    // and proceeds to set pinState to 1.
    if ( myDataOut & (1<<i) ) {
      pinState= 1;
    }
    else {  
      pinState= 0;
    }

    //Sets the pin to HIGH or LOW depending on pinState
    digitalWrite(myDataPin, pinState);
    //register shifts bits on upstroke of clock pin  
    digitalWrite(myClockPin, 1);
    //zero the data pin after shift to prevent bleed through
    digitalWrite(myDataPin, 0);
  }

  //stop shifting
  digitalWrite(myClockPin, 0);
}


//blinks the whole register based on the number of times you want to 
//blink "n" and the pause between them "d"
//starts with a moment of darkness to make sure the first blink
//has its full visual effect.
void blinkAll_2Bytes(int n, int d) {
  digitalWrite(latchPin, 0);
  shiftOut(dataPin, clockPin, 0);
  shiftOut(dataPin, clockPin, 0);
  digitalWrite(latchPin, 1);
  delay(200);
  for (int x = 0; x < n; x++) {
    digitalWrite(latchPin, 0);
    shiftOut(dataPin, clockPin, 255);
    shiftOut(dataPin, clockPin, 255);
    digitalWrite(latchPin, 1);
    delay(d);
    digitalWrite(latchPin, 0);
    shiftOut(dataPin, clockPin, 0);
    shiftOut(dataPin, clockPin, 0);
    digitalWrite(latchPin, 1);
    delay(d);
  }
}

How would I adjust this so that I can use an extra pin 3 and 5 in the same Array, so I can control them all as one group?

What do you mean, "the extra pin 3 and 5"? What is special about 3 and 5? Pins on what?

If Pins 3 and 5 are just a couple of spare pins that you want to integrate into the beginning of your LED display, you could do a trick like this example (untested). This is the first part of your loop. You’ll also have to declare the pins as output in setup() and fix blinkAll(). Better would be to chain in a second shift register.

void loop() {
  for (int j = 0; j <= 10; j++) {
    if ( j >= 0 &&  j <= 8 ) {
      digitalWrite( 3, HIGH ) ;
      digitalWrite( 5, HIGH) ;
      //load the light sequence you want from array
      data = dataArray[j];
      //ground latchPin and hold low for as long as you are transmitting
      digitalWrite(latchPin, 0);
      //move 'em out
      shiftOut(dataPin, clockPin, data);
      //return the latch pin high to signal chip that it
      //no longer needs to listen for information
      digitalWrite(latchPin, 1);
    }
    else if (  j == 9 ) {
      digitalWrite( 3, HIGH ) ;
      digitalWrite( 5, LOW ) ;
    }
    else if (  j == 10 ) {
      digitalWrite( 3, LOW ) ;
      digitalWrite( 5, LOW ) ;
    }
    else { }
    Serial.println("No. 1");
    Serial.println(j);
    delay(300);
  }
}

Or just add a 2nd daisy chained shift register. Make these int dataArray[0] = 0xFF; //0b11111111 and shiftOut the highByte() and lowByte() to the registers.