A spring of spring constant 90 N/m holds a block of 0.40 kg on a smooth floor…

(a) $T_0 = 2\pi\sqrt{\dfrac{0.40\ \text{kg}}{90\ \text{N/m}}} = 0.419\ \text{s}$.

(b) The shorter piece is $\tfrac13$ of the length, so $k_1 = 3 \times 90\ \text{N/m} = 270\ \text{N/m}$. The longer piece is $\tfrac23$ of it, so $k_2 = \tfrac32 \times 90\ \text{N/m} = 135\ \text{N/m}$.

(c) Move the block by $x$. One piece stretches and the other is squeezed, and both push it back. So $k = 270\ \text{N/m} + 135\ \text{N/m} = 405\ \text{N/m}$, and $T = 2\pi\sqrt{\dfrac{0.40\ \text{kg}}{405\ \text{N/m}}} = 0.197\ \text{s}$. The ratio is $\sqrt{\dfrac{90}{405}} = 0.471$.