Fourier transform of the regular tempered distributions induced by $x^\alpha$, $\sin(\langle a, x \rangle_{\mathbb R^n})$, and $\mathbb 1_{[-1,1]}$

1. Since $(-ix)^\alpha\hat f = \widehat{D^\alpha f},$
   $$ \int_{\mathbb{R^n}} x^\alpha \mathcal F\phi dx = i^{|\alpha|}\int_{\mathbb{R^n}}\mathcal F(D^\alpha\phi) dx = (2\pi)^{n/2}i^{|\alpha|}\mathcal F(\mathcal F(D^\alpha\phi))(0) = (2\pi)^{n/2}i^{|\alpha|}D^\alpha\phi(0). $$

2. First note $$ \int_{\mathbb{R^n}} e^{i \langle a, x \rangle} \mathcal F\phi(x) dx = (2\pi)^{n/2}\phi(a). $$ Now since $2i\sin(t) = e^{it} -e^{-it}$ we get $$\int\sin(\langle a, x \rangle) \mathcal F\phi (x) dx = (2\pi)^{n/2}\frac{ \phi(a) - \phi (-a)}{2i}.$$

3. Yes, the fourier transform of a distribution induced by an $L^1$ function $f$ is that induced by the fourier transform of $f$, and $$ \mathcal F(\mathbb1_{[-1,1]})(x) = \frac1{\sqrt{2\pi}} \frac{2\sin(x)}x.$$

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(convention for Fourier Transform) here, we use the Fourier/Inverse Fourier transform pair mentioned in comments $$ \mathcal Ff(\xi) = (2\pi)^{-n/2}\int f(x) e^{-i\xi\cdot x} dx, \quad \mathcal F^{-1}f(\xi) = (2\pi)^{-n/2}\int f(x) e^{i\xi \cdot x} dx = \mathcal Ff(-\xi)$$ Wolfram Mathworld has helpfully laid out the different conventions of the Fourier Transform here.