As $\theta \to 0^{+}$, $x^{2/\theta }\to \infty$ when $|x|>1$ and $x^{2/\theta }\to 0$ when $|x|<1$. So $f$ is defined piecewise around $x=1$ and $x=-1$.

At $x=1$:

For $x\to 1^{-}$ (so $|x|<1$, the $x^{2/\theta }$ terms vanish): $f(x)\to \cos \pi x\to \cos \pi =-1$.

For $x\to 1^{+}$ (so $|x|>1$, dividing numerator and denominator by $x^{2/\theta }$): $f(x)\to \frac{-\sin (x-1)}{x-1}\to -1$.

At $x=1$: substitute directly $\to \cos \pi =-1$. Thus $f$ is continuous at $x=1$, so (I) is false.

At $x=-1$:

For $x\to -1^{-}$ (so $|x|>1$): $f(x)\to \frac{-\sin (x-1)}{x-1}\to \frac{-\sin (-2)}{-2}=\frac{\sin 2}{-2}$.

For $x\to -1^{+}$ (so $|x|<1$): $f(x)\to \cos \pi x\to \cos (-\pi )=-1$.

The two side limits differ, so $f$ is discontinuous at $x=-1$, making (II) false as well.

$\therefore$ Neither (I) nor (II) is True.