Answer
a) $2\;\rm Hz$
b) $1.2\;\rm cm$
Work Step by Step
a) We have here a system that undergoes a simple harmonic motion.
And we know that the frequency of such a system is independent of the amplitude and the maximum velocity since it is given by
$$\omega=2\pi f=\sqrt{\dfrac{k}{m}}$$
$$f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}$$
We can see that the right side is constant, so the frequency remains the same after the force blowing is finished.
$$f'=f=\color{red}{\bf 2}\;\rm Hz$$
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b)
Before applying the thrust force, the maximum speed of the block is given by the conservation of energy.
$$\color{red}{\bf\not} \frac{1}{2} mv_{1,\rm max}^2=\color{red}{\bf\not} \frac{1}{2} kA_1^2$$
So,
$$mv_{1,\rm max}^2= kA_1^2$$
Noting that we used the maximum velocity since the force thrust applied on the block when it passes through the equilibrium point at which the speed is maximum.
Thus,
$$ v_{1,\rm max}^2 =\dfrac{ k}{m}A_1^2$$
Taking the square root,
$$ v_{1,\rm max} =\sqrt{\dfrac{ k}{m}}A_1 $$
where $\sqrt{\dfrac{ k}{m}}=\omega=2\pi f$
$$ v_{1,\rm max} =2\pi f A_1 \tag 1$$
Noting that $\omega$ and $f$ remain constant before and after applying the thrust force.
We know that the force provides an impulse on the block, so
$$J_x=\Delta p=F\Delta t$$
Hence,
$$mv_{2,\rm max}-mv_{2,\rm max}=F\Delta t$$
Plugging from (1);
$$mv_{2,\rm max}-m(2\pi f A_1)=F\Delta t$$
We know that the velocity of the block at the equilibrium point is given by $A\omega$, so $v_{2,\rm max}=\omega A_2=2\pi f A_2$.
$$2\pi f A_2m - 2\pi f A_1m =F\Delta t$$
Solving for $A_2$;
$$ A_2 =\dfrac{F\Delta t+2\pi f A_1m}{2\pi f m} =\dfrac{F\Delta t}{2\pi f m}+ A_1 $$
Plugging the known, and recall that the force direction is toward the left while the block's initial velocity was to the right.
$$ A_2 =\dfrac{(-20) (1\times 10^{-3})}{2\pi (2) (0.2)}+ 0.02=\bf 0.0120\;\rm m$$
$$A_2 =\color{red}{\bf 1.20}\;\rm cm$$