15-79.

What we have to note here is that the condition of the cars before the bottom one breaks away enables us to determine the spring constant of the rope since the system was in static equilibrium. The component of weight along the incline of the 3 cars balances the spring force, so Newton's Second Law gives us

$$3mg\sin(\theta)=F_s=kx$$

where $m$ is the mass of one car and $x$ the stretch of the spring, so solving for $k$ yields

$$k=3mg\sin(\theta)/x=9.81\times 10^5\;{\rm N/m},$$

which allows us to determine the frequency. But, be careful, only two of the cars oscillate, so the mass now is $2m$!

Now, for the amplitude we note that the breaking free of the bottom car set our initial conditions: the string is stretched by $x=15\;{\rm cm}$ and the initial velocity is zero. When the bottom one breaks free our forces are unbalanced so the net force up the incline is still $f_s=3mg\sin(\theta)$, but the gravitation part downward is only $2mg\sin(\theta)$. The net force is then $kA$ (since we oscillate about the new equilibrium position), so

$$A=\frac{F_{net}}{k}=\frac{mg\sin(\theta)}{3mg\sin(\theta)/x}=\frac{x}{3}.$$

Another way to do this part is to to write $A=x-\Delta l$, and find the stretch of the spring from its unstretched length as we did before. Both expressions should agree.