Answer
$A = 0.11~m$
$T = 1.7~s$
Work Step by Step
We can use conservation of momentum to find the speed of the two gliders just after the collision.
$(m_1+m_2)v_2 = m_1v_1$
$v_2 = \frac{m_1v_1}{m_1+m_2}$
$v_2 = \frac{(0.25~kg)(1.20~m/s)}{0.25~kg+0.50~kg}$
$v_2 = 0.40~m/s$
Let $M$ be the the total mass of the two gliders. We can find the amplitude.
$\frac{1}{2}kA^2 = \frac{1}{2}Mv_{max}^2$
$A = \sqrt{\frac{M}{k}}~v_{max}$
$A = \sqrt{\frac{0.75~kg}{10~N/m}}~(0.40~m/s)$
$A = 0.11~m$
We then find the period:
$T = 2\pi~\sqrt{\frac{M}{k}}$
$T = 2\pi~\sqrt{\frac{0.75~kg}{10~N/m}}$
$T = 1.7~s$
