Answer
a) $f=2.66Hz$
b) $A=3.5cm$
Work Step by Step
a) The frequency of the motion can be calculated by $$f=\frac{\omega}{2\pi}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$
The spring constant $k=112N/m$ and the block's mass $m=0.4kg$, so $$f=2.66Hz$$
b) At the top, the spring reaches its maximum downward acceleration $a_{max}$, while the block is under gravitational acceleration $g$. The block will lose contact with the spring when this $a_{max}\gt g$, since the spring will go down faster than the block.
Therefore, at the point when the block loses contact with the spring, $$a_{max}=g$$ $$A\omega^2=A(2\pi f)^2=g$$ $$A=\frac{g}{(2\pi f)^2}=3.5\times10^{-2}m=3.5cm$$
