2. $sin(n \pi) = 0$, so you require $x^2 + 1 = n \pi$. This occurs at $x = \sqrt{n \pi -1}$, in this case $n = 2,3,4,5$.
When $sin(x^2+1)=0$, the angle must be some multiple of $\pi$. So $x^2+1=n\pi$ would be a solution.