Answer
The oscillation frequency when no one is sitting on the horse is $2.5~Hz$
Work Step by Step
We can find the spring constant $k$:
$kx = mg$
$k = \frac{mg}{x}$
$k = \frac{(24~kg)(9.80~m/s^2)}{0.28~m}$
$k = 840~N/m$
Let $m_c$ be the mass of the child. We can find the mass of the horse $m_h$:
$\omega = 2\pi~f$
$\sqrt{\frac{k}{m_c+m_h}} = 2\pi~f$
$m_c+m_h = \frac{k}{(2\pi~f)^2}$
$m_h = \frac{k}{(2\pi~f)^2}-m_c$
$m_h = \frac{840~N/m}{(2\pi)^2(0.88~Hz)^2}-(24~kg)$
$m_h = 3.476~kg$
We can find the oscillation frequency when no one is sitting on the horse:
$2\pi~f = \omega$
$f = \frac{\omega}{2\pi}$
$f = \frac{1}{2\pi}~\sqrt{\frac{k}{m_h}}$
$f = \frac{1}{2\pi}~\sqrt{\frac{840~N/m}{3.476~kg}}$
$f = 2.5~Hz$
The oscillation frequency when no one is sitting on the horse is $2.5~Hz$.
