Stirchartz's book ("A guide to distribution theory and fourier transforms" ) has Chapter 1 exercises
Here $\mathcal{D(\mathbb{R}^1)}$ is a set of test functions $\phi:\mathbb{R} \rightarrow \mathbb{R}$ with a compact support away from the boundaries of $\mathbb{R}$
3) For any a>0, show that <$f_a$,$\phi$> = $\int_{-\infty}^{-a}\frac{\phi(x)}{|x|}dx + \int_{a}^{\infty} \frac{\phi(x)}{|x|}dx + \int_{-a}^{a}\frac{\phi(x)-\phi(0)}{|x|}dx $ is a distribution.
4) Show that = $\int_{-\infty}^{\infty} \frac{\phi(x)}{|x|}$ for any $\phi\in \mathcal{D(\mathbb{R}^1)}$ for which $\phi(0) = 0$.
One uses mean value theorem on test function $\phi$, but that only proves finiteness not convergence. Infact isn't exercise 4 a special case of 3?
For (3) we need to estimate the sum of those three integrals in terms of uniform norms $\|\phi\|$, $\|\phi'\|$, etc. Let $supp(\phi)\subset[-R,R]$, $R>0$. We write
$$\left|\int_{-\infty}^{-a}\frac{\phi(x)}{|x|}dx\right|\begin{cases}=0,&\text{if }a\ge R,\\ \le \|\phi\|\int_{-R}^{-a}\frac{dx}{|x|}=\|\phi\|\ln(R/a),&\text{if }R>a.\end{cases}$$
The similar estimation can be made for the integral $\int_{a}^{\infty}\frac{\phi(x)}{|x|}dx$.
Now we estimate
$$\left|\int_{-a}^a\frac{\phi(x)-\phi(0)}{|x|}dx\right|=\left|\int_{-a}^0\frac{\phi(x)-\phi(0)}{|x|}dx+\int_{0}^a\frac{\phi(x)-\phi(0)}{|x|}dx\right|=$$ $$=\left|-\int_{-a}^0\frac{\phi'(0)x+\phi''(\xi_x)x^2/2}{x}dx+\int_{0}^a\frac{\phi'(0)x+\phi''(\xi_x)x^2/2}{ x }dx\right|\le \frac 12\|\phi''\|a^2.$$
It seems that you forgot to enter some details of the part (4), so I can't help with that part.