#### Answer

a) $A=3.59\times10^{-2}m$ and $f=4.24Hz$
b) $A=5.08\times10^{-2}m$ and $f=3Hz$

#### Work Step by Step

a) When the block is split at $v_{max}$, the remaining part still retains this maximum speed $v_{max}$. However, because the block is half split, its mass reduces by half, and $KE_{max}=\frac{1}{2}mv_{max}^2$ will reduce by half, too.
Because of energy conservation, $KE_{max}=PE_{elastic, max}$. Therefore, the maximum elastic energy will reduce by half. $PE_{elastic}$ is proportional with amplitude in the form of $A^2$, so as $PE_{elastic}$ reduces by 2, $A$ reduces by $\sqrt2$. So, $$A=\frac{5.08\times10^{-2}m}{\sqrt2}=3.59\times10^{-2}m$$
We know $v_{max}=A\omega=2\pi Af$. While $v_{max}$ remains the same, $A$ decreases $\sqrt2$ times, so $f$ will increase $\sqrt2$ times: $$f=sqrt2\times3=4.24Hz$$
b) At one of the extreme positions, there is no kinetic energy, only elastic energy $PE_{elastic}=\frac{1}{2}kx^2$. Both spring constant $k$ and the spring's displacement $x$ at these extreme positions are not affected by the block being halved, so $PE_{elastic}$ remains the same.
Therefore, $A$ and $f$ also remain the same.