Answer
a) $2\;\rm rad/s$
b) $15\;\rm cm$
Work Step by Step
We know that at $t=0$, the block is at the maximum distance from the equilibrium point, which we call the amplitude $x=A$.
And we know that the position of an object that undergoes a simple harmonic motion is given by
$$x_{(t)}=A\cos(\omega t+\phi_0)$$
So, at $t=0$, $$x_{(0)}=A\tag 1$$
which means that $\phi_0=0$
Thus,
$$x_{(t)}=A\cos(\omega t)\tag2$$
And at $t=0.685$ s, $x=3$ cm
$$3=A\cos(0.685\omega)\tag a$$
and at $t=0.886$ s, $x=-3$ cm
$$-3=A\cos(0.886\omega)\tag b$$
From (a) and (b);
$$ \color{red}{\bf\not}A\cos(0.685\omega)=- \color{red}{\bf\not}A\cos(0.886\omega)$$
$$ \cos(0.685\omega)=- \cos(0.886\omega)\tag c$$
From the hint given by the author, $- \cos(0.886\omega)= \cos(\pi-0.886\omega)$
Plugging into (c);
$$ \cos(0.685\omega)= \cos(\pi-0.886\omega)$$
Hence,
$$ 0.685\omega= \pi-0.886\omega$$
$$ 0.685\omega+0.886\omega= \pi$$
$$ \omega= \dfrac{\pi}{1.571}=1.9999\;\rm rad/s$$
$$ \omega= \color{red}{\bf 2.0}\;\rm rad/s$$
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b) Plugging the final result above into (a);
$$3=A\cos(0.685\times 2) $$
Hence,
$$A=\dfrac{3}{\cos(1.37)}= \color{red}{\bf15}\;\rm cm$$
