The first problem is correct.

For the second, consider

tan(theta) = sqrt{3}/3

What angle (theta) is this? (theta) = (pi)/6 rad.

But we need (theta) for tan(theta) = -sqrt{3}/3. The reference angle is (pi)/6 rad and tangent is negative in QII. So (theta) = (pi) - (pi)/6 = 5(pi)/6 rad.

-Dan