Answer
a) $0\;\rm N$
b) $-222\;\rm N$
Work Step by Step
We know, from Newton's second law, that the net force is given by
$$\sum F=ma$$
And since the only force exerted on the astronaut now is the spring force,
$$F_{sp}=ma\tag 1$$
Now we need to find the acceleration of the astronaut by differentiating the given equation twice.
We know that the position formula is given by
$$x=0.3\sin(\pi t)$$
Thus, the speed is given by
$$\dfrac{d}{dt}x=\dfrac{d}{dt}(0.3\sin(\pi t))$$
$$v =0.3\pi \cos(\pi t)$$
Hence, the acceleration is given by
$$\dfrac{d}{dt}v=\dfrac{d}{dt}(0.3\pi \cos(\pi t))$$
$$a =-0.3\pi^2 \sin(\pi t)$$
Plugging into (1);
$$F_{sp}=m \left[-0.3\pi^2 \sin(\pi t)\right]$$
$$F_{sp}= -0.3m\pi^2 \sin(\pi t) $$
a) At $t=1$ s:
$$F_{sp}= -0.3\cdot 75\pi^2 \sin(\pi \cdot 1)=\color{red}{\bf 0}\;\rm N $$
a) At $t=1.5$ s:
$$F_{sp}= -0.3\cdot 75\pi^2 \sin(\pi \cdot 1.5)=\color{red}{\bf -222}\;\rm N $$
