1. We will prove that $\displaystyle \mathbb{R}\backslash \operatorname{supp} u\supset \mathbb{R}\backslash Z(f)$. According to the definition, for any $\displaystyle x\in \mathbb{R}\backslash \operatorname{supp}u$, there exists a neighborhood $\displaystyle V_{x}$ of $x$ satisfying $\displaystyle \left\langle u,\varphi \right\rangle=0$ for any $\displaystyle \varphi\in C_{0}^{\infty}(V_{x})$. And for any $\displaystyle x\in \mathbb{R}\backslash Z(f)$, we have $\displaystyle f(t)\neq0$ for $\displaystyle t$ in a neighborhood of $x$. We can assume that it's $\displaystyle V_x$. Now we know, for any $\displaystyle \varphi\in C_{0}^{\infty}(V_{x})$, we have $\displaystyle \left\langle u,\varphi \right\rangle=\left\langle u\cdot f,\frac{\varphi}{f} \right\rangle=0$ because of $\displaystyle \frac{\varphi}{f}\in C_{0}^{\infty}(V_{x})$ and $\displaystyle f\neq0$ in that set. So we get $\displaystyle \mathbb{R}\backslash \operatorname{supp} u\supset \mathbb{R}\backslash Z(f)$.

2. MSE link If $|c_{n}|\le C(1+n^{p})$, then we have $\displaystyle \left|\left\langle \sum_{n=1}^{\infty}c_{n}\delta_{n},\varphi \right\rangle\right|\le \sum_{n=1}^{\infty}\left|c_{n}\right|\left|\left\langle \delta_{n},\varphi \right\rangle\right|\le C\sum_{n=1}^{\infty}(1+n)^{p}\left|\varphi(n)\right|\le C\sup\left|(1+|x|)^{p+2}\varphi(x)\right|\sum_{n=1}^{\infty}(1+n)^{-2}=C'\sup\left|(1+\left|x\right|)^{p+2}\varphi(x)\right|$. Then we know it's a tempered distribution.
   If it's known that $\displaystyle \sum_{n=1}^{\infty}c_{n}\delta_n$ is a tempered distribution and $|c_{n}|$ cannot be dominated by a polynomial, then we will show the contradiction. Take an increasing subsequence $\{c_{x_{n}}\}$ of $\{c_n\}$ which satisfying $\displaystyle c_{x_{n}}>x_{n}^{n}$ and let $\displaystyle d_{k}=\begin{cases}x_{n}^{-n}&k=x_{n}\\0&k\neq x_{n}\end{cases}$, then we have $\displaystyle \sum_{n=1}^{\infty}c_{n}d_{n}$ diverges. Notice that $\displaystyle \varphi=\sum_{n=1}^{\infty}\psi_{n}$ satisfies $\displaystyle \varphi\in C^{\infty}(\mathbb{R})$, where $\displaystyle \psi_{n}(x)=\begin{cases}d_{n}e^{4+\frac{1}{\left(x-\frac{2n-1}{2}\right)\left(x-\frac{2n+1}{2}\right)}}&x\in\left(\frac{2n-1}{2},\frac{2n+1}{2}\right)\\0&x\notin\left(\frac{2n-1}{2},\frac{2n+1}{2}\right)\end{cases}$. And $\displaystyle \varphi\in \mathcal{S}(\mathbb{R})$ because $\displaystyle d_{n}\le \frac{1}{n^{k}}$ for any $\displaystyle k$ when $\displaystyle n\to +\infty$. Finally, we can get contradiction by $\displaystyle \left\langle \sum_{n=1}^{\infty}c_{n}\delta_{n},\varphi \right\rangle=+\infty$.

3. According to the Fourier transform table, we know $\displaystyle \mathcal{F}\left(\pi e^{-2\pi|x|}\right)=\frac{1}{1+\xi^{2}}$. So we have $\displaystyle \mathcal{F}\left(\frac{1}{1+x^{2}}\right)=\mathcal{F}\left(\mathcal{F}\left(\pi e^{-2\pi|x|}\right)\right)=\pi e^{-2\pi |\xi|}$.

4. We know $\displaystyle \mathcal{F}(e^{-|x|})=\frac{2}{4\pi^{2}\xi^{2}+1}$ from Fourier transform table immediately.

5. We know $\displaystyle \mathcal{F}(\mathbf{1}_{x\ge0})=\frac{1}{2}\left(\frac{1}{i\pi\xi}+\delta_{0}(\xi)\right)$ and $\displaystyle \mathcal{F}(f(x-a))=e^{-2\pi i \xi a}\mathcal{F}(f(x))$, so we have $\displaystyle \mathcal{F}(\mathbf{1}_{x\ge a})=\frac{e^{-2\pi i\xi a}}{2}\left(\frac{1}{i\pi\xi}+\delta_{0}(\xi)\right)$.

6. First, we know $\displaystyle \frac{1}{x+i0}=\text{p.v.}\frac{1}{x}-i\pi\delta_{0}$. Then we know $\displaystyle \mathcal{F}\left(\frac{1}{x+i0}\right)=\mathcal{F}\left(\text{p.v.}\frac{1}{x}\right)-\mathcal{F}(i\pi\delta_{0})=-i\pi\text{sign}(\xi)-i\pi$.

7. Notice that $\displaystyle \mathcal{F}(H(x))=\frac{1}{2\pi i\xi}+\frac{\delta_{0}}{2}$, so we know $\displaystyle \mathcal{F}(x^{k}\mathbf{1}_{x\ge0})=\frac{k!}{(2\pi i\xi)^{k+1}}+\frac{\delta_{0}^{(k)}}{2(-2\pi i)^{k}}$ from $\displaystyle \mathcal{F}(xu)=-\frac{\partial_{x}\mathcal{F}(u)}{2\pi i}$. Now we have $\displaystyle \mathcal{F}(P(x)\mathbf{1}_{x\ge 0})=\sum_{k=0}^{n}a_{k}\left(\frac{k!}{(2\pi i\xi)^{k+1}}+\frac{\delta_{0}^{(k)}}{2(-2\pi i)^{k}}\right)$ for $\displaystyle P(x)=\sum_{k=0}^{n}a_{k}x^{k}$.

8. We know $\displaystyle \frac{1}{(x-i0)^{\alpha+1}}=\begin{cases}x_{+}^{-\alpha-1}+e^{\pi i(\alpha+1)}x_{-}^{-\alpha-1}&\alpha\notin\mathbb{N}\\x_{+}^{-\alpha-1}+(-1)^{\alpha+1}x_{-}^{-\alpha-1}-\frac{\pi i(-1)^{\alpha+1}\delta_{0}^{(\alpha)}}{\alpha!}&\alpha\in \mathbb{N}\end{cases}$, where $\displaystyle x_{-}^{a}=\begin{cases}0&x\ge0\\|x|^{a}&x<0\end{cases}$, by the representation of $x_{+}^{-k}$ in sense of distribution. Notice that $\displaystyle \int_{0}^{+\infty}x^{\alpha}e^{-2\pi i x\xi}\mathrm{d}x\xlongequal{t=\varepsilon+i\xi}\lim_{\varepsilon\to0^{+}}\int_{0}^{+\infty}x^{\alpha}e^{-2\pi tx}\mathrm{d}x\xlongequal{u=2\pi tx}\lim_{\varepsilon\to0^{+}}\int_{0}^{+\infty}\frac{u^{\alpha}}{(2\pi t)^{\alpha}}e^{-u}\frac{\mathrm{d}u}{2\pi t}=\lim_{\varepsilon\to0^{+}}\frac{1}{(2\pi t)^{\alpha+1}}\int_{0}^{+\infty}u^{\alpha}e^{-u}\mathrm{d}u=\lim_{\varepsilon\to0^{+}}\frac{\Gamma(\alpha+1)}{(2\pi t)^{\alpha+1}}=\Gamma(\alpha+1)\left(-\frac{i}{2\pi}\right)^{\alpha+1}\frac{1}{(x-i0)^{\alpha+1}}$, then we know the representation of $\displaystyle \mathcal{F}(\operatorname{pf} x_{+}^{\alpha})$.

9. According to Problem 8, we know $\displaystyle \mathcal{F}(\operatorname{pf}x_{+}^{\alpha})=\Gamma(\alpha+1)\left(-\frac{i}{2\pi}\right)^{\alpha+1}\frac{1}{(x-i0)^{\alpha+1}}=\Gamma(a+1)\frac{x_{+}^{-\alpha-1}+e^{\pi i(\alpha+1)}x_{-}^{-\alpha-1}}{(2\pi i)^{\alpha+1}}$, where $\displaystyle x_{-}^{a}=\begin{cases}0&x\ge0\\|x|^{a}&x<0\end{cases}$.

10. We have $\displaystyle \mathcal{F}\left(e^{\alpha x^{2}}\right)=\sqrt{\frac{\pi}{\alpha}}e^{-\frac{(\pi\xi)^{2}}{\alpha}}$ by Fourier transform table. So we get $\displaystyle \mathcal{F}\left(e^{-ix^{2}}\right)=\sqrt{\pi i}e^{-i\pi^{2}\xi^{2}}$.