1.) Let $x_0 \in \mathbb{R}$. Recall that $\sin \alpha - \sin \beta = 2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$. Then as $x\rightarrow x_0$ we have: \begin{align*} \left|f(x) - f(x_0)\right| &= \left|\sin x^2-\sin x_0^2\right|\\ &= \left|2 \cos \frac{x^2+x_0^2}{2} \sin \frac{x^2-x_0^2}{2}\right|\\ &\le 2 \left|\sin \frac{x^2-x_0^2}{2}\right| \rightarrow 0. \end{align*} Just choose your $\delta$ so that the last term is small enough. If you can cite the continuity of $\sin$ and $x^2,$ that's simple. Otherwise, you might have to go through their proofs as well.
2.) No, $f$ is not uniformly continuous on $\mathbb{R}$. We can, for any $\delta$, choose $x_0$ so large that $x_0^2 = n\pi$ for some $n\in \mathbb{N}$, $x^2 = (n +\frac{1}{2}) \pi$, and $|x-x_0|<\delta$ (this might also require some additional proof). But then $|f(x) - f(x_0)| = 1$, so $f$ cannot be uniformly continuous.
You can fill in the details, but that's a general approach.
Edit: the other responses are much cleaner, but feel free to get your hands dirty with this one. ;)