Answer
(a) less than
(b) I
Work Step by Step
(a) We know that $f=\frac{1}{2\pi}(\sqrt{\frac{K}{m}})$. This equation shows that with an increase of mass 'm', the frequency decreases. Thus, the resulting frequency of oscillation is less than what it was before.
(b) We know that the correct option is statement (I) -- that is, increasing the mass on a spring increases its period, and hence decreases its frequency.
