Rather than answering your specific question I believe it's better to explain simple harmonic motion. We say some dynamical variable [tex]x[tex] is in simple harmonic motion if its value at time t is [tex]x(t)=A\cos(\omega t+\phi)[tex] for some constants [tex]A[tex] and [tex]\phi[tex]. The rate of change of [tex]x[tex] (in your case this is the velocity) as a function of time can be found by taking the derivative. In this case, [tex]v(t)=x'(t)=-A\omega\sin(\omega t + \phi)[tex]. The rate of change of the rate of change (in your case acceleration) can once again be found by taking the derivative [tex]a(t)=v'(t)=-A\omega^2\cos(\omega t +\phi)[tex]. The phase of the motion is just the argument of the cosine function, that is, [tex]\omega t+\phi[tex]. The frequency is the number of cycles which the motion makes in 1 second. This is of course just the inverse of the time it takes the oscillator in do one cycle. Since the cosine has period [tex]2\pi[tex] the time it take to make one cycle is [tex] T = \frac{2\pi}{\omega}[tex]. Therefore, the frequency is [tex]f=\frac{\omega}{2\pi}[tex].
In your problem [tex] A=7.1m[tex], [tex]\omega=5\pi rad/s[tex] and [tex]\phi=\pi/5rad[tex]. You've just gotta replace this values in the equations above to solve for what the problem asks.