Here is my solution for exercise #2 using recursion and pattern matching.. Easy to understand

parenthesis

def parenthesis(input: List[Char], count: Int):Boolean ={
  input match {
    case Nil if count==0 => true
    case Nil if count!=0 => false
    case head :: tail if count==(-1) => false
    case head :: tail if head == '('=> parenthesis(tail, count + 1)
    case head :: tail if head == ')'=> parenthesis(tail, count - 1)
    case head :: tail if head!='(' && head!=')' => parenthesis(tail, count)
  }
}