Dirac delta solutions

First, if $f_1$ and $f_2$ are solutions to $Tf=g,$ where $T$ is some linear operator and $g$ is given, then $f_1-f_2$ is a solution to $Tf=0.$ Therefore we will study $x f(x)=0.$ It can easily be shown in distribution theory that $x\delta(x)=0,$ $x^2\delta(x)=0,$ and $x^2\delta'(x)=0,$ but since you're studying Fourier transforms I will give an explanation using Fourier transforms:

Take the equation $x f(x) = 0$ and apply the Fourier transform to both sides. You get $i\hat{f}'(\xi) = 0.$ This is a differential equation with solutions $\hat{f}(\xi) = C,$ where $C$ is a constant. Taking the inverse Fourier transform gives us $f(x) = C\delta(x).$

Likewise, $x^2 f(x) = 0$ transforms to $-\hat{f}''(\xi)=0$ with solutions $\hat{f}(\xi) = A\xi+B,$ i.e. $f(x) = -iA\delta'(x)+B\delta(x).$

So, informally, the Dirac $\delta$ is zero everywhere except at $0$ and has integral $1$. So, informally, $\delta$ is infinite at $0$, therefore $\delta$ is not admitted by traditional analysis. In regular analysis, given $x.f(x) = a$, we divide both sides by $x$ to obtain $f(x) = a/x$ but we can add any number (say $b$) times $\delta(x)$ on to $a/x$ as when $x$ is not zero any number times $\delta(x)$ is just $0$ and when $x$ is $0$ then $x.f(x)$ is still zero and so adding $b. \delta (x)$ on to $a/x$ does not change the truth of the fact that $x.f(x)=a$.

Now perhaps the other solution will make sense but it might help to know that if $\delta^{'}(x)$ is the derivative of the Dirac function then $\delta^{'}(x)=-\delta(x)/x$ so $\delta^{'}$ is 'even more infinite' than $\delta(x)$ :).

To complete the good answer given by md2perpe, you just need to get one particular solution of the equations. In dimension $1$ however, $1/x$ and $1/x^2$ are not locally integrable functions, and so one should define them as principal values (and one sometimes writes $\mathrm{P}(\tfrac{1}{x}) = \mathrm{pv.}(\tfrac{1}{x})$ and $\mathrm{fp.}(\tfrac{1}{x^2})$ for principal value and finite part). For any smooth and compactly supported function $\varphi$, they are defined by $$ \langle\mathrm{P}(\tfrac{1}{x}),\varphi\rangle = \int_{\mathbb{R}} \frac{\varphi(x)-\varphi(0)}{x}\,\mathrm{d} x $$ which can also be written $\langle\mathrm{P}(\tfrac{1}{x}),\varphi\rangle = \lim_{\varepsilon\to 0}\int_{|x|>\varepsilon} \frac{\varphi(x)}{x}\,\mathrm{d} x$. One can easily verify that $$ x\, \mathrm{P}(\tfrac{1}{x}) = 1 $$

So the general solution for the first equation is $$ f(x) = a \, \mathrm{P}(\tfrac{1}{x}) + b \, \delta_0 $$

In the same way, one can define $$ \langle\mathrm{pf.}(\tfrac{1}{x^2}),\varphi\rangle = \int_{\mathbb{R}} \frac{\varphi(x)-\varphi(0)- x \varphi'(0)}{x^2}\,\mathrm{d} x $$ and then the general solution of the second equation is $$ f(x) = a \, \mathrm{pf.}(\tfrac{1}{x^2}) + b \, \delta_0 + c \, \delta_0' $$

Edit: $\delta_0(x)/x$ has no clear meaning in distribution theory. However, as indicated by Simon Terrington, one could define $\delta_0(x)/x = -\delta_0'(x)$ since it is one of the solution of the equation $$ x\,g(x) = -\delta_0(x). $$ The general solution being $g(x) = -\delta_0' + c\, \delta_0$. It is better to use just $\delta_0'$.