Since we're only thinking in terms of the first ball, this problem essentially tells us the future; we know in advance that the first ball couldn't be 4, because the second ball is, and none of the balls were replaced after drawing. This leaves us with 9 balls, so all of our probabilities will be out of 9.
There are 3 numbers between 1-10 that are less than 4 - 1, 2, and 3 - so the chance of picking one on the first draw is 3/9, or 1/3.
With all 10 balls in play, we'd have 5 different even numbers - 2, 4, 6, 8, and 10 - but since we've removed 4 from the running, we only have 4 even numbers out of the 9 balls in play, giving us a 4/9 chance of drawing an even number.