The problems are:
Show that for each $\phi \in C_{C}^{\infty}(\mathbb{R})$, the principal value integral $$\text{p.v.}\int\frac{\phi(x)}{x}\text{d}x = \lim_{\epsilon \to 0^+}(\int_{(-\infty,-\epsilon)}\frac{\phi(x)}{x}\text{d}x+\int_{(\epsilon,\infty)}\frac{\phi(x)}{x}\text{d}x)$$ exists and is a distribution. What is its order?
Find a distribution $u \in \mathcal{D}'(\mathbb{R})$ such that $u=1/x$ on $(0,\infty)$ and $u=0$ on $(-\infty,0)$.
So far I think I can show that given $\phi \in C_{C}^{\infty}(\mathbb{R})$ and the function $f$ defined by $$f: (0,\infty) \to \mathbb{C}$$ $$x \mapsto \frac{\phi(x)-\phi(-x)}{x},$$ we have that $f$ is continuous and $\int_{[0,1]}f$ exists. Furthermore, I can show that, by the Mean Value Theorem, $|\int_{[0,1]}f| \leq 2\sup_{x \in \mathbb{R}}|\phi'(x)|$, $|\int_{[1,\infty)}f\text|<2\sup_{x\in\mathbb{R}}|x\phi(x)|$.
This then means the principal value integral is (obviously) linear and bounded by the appropriate semi-norm, and is therefore a distribution, correct? The inequalities above tell me it's order is then $1$, correct? This then settles the first question, correct?
(The above is based on: https://en.wikipedia.org/wiki/Cauchy_principal_value#Well-definedness_as_a_distribution)
Now, for question 2, I'm supposed to find a function $u: C_C^{\infty}(\mathbb{R}) \to \mathbb{C}$, correct? I am then having trouble understanding what it means to say that "$u=1/x$ on $(0,\infty)$ and $u=0$ on $(-\infty,0)$", given the that the domain of $u$ is not (a subset of) the real line. Anyway, I think that this entire problem has something to do with the fact that $(\log \circ |.|)'=\frac{1}{.}$, correct?
For your question 2 (it's nice when a question only has one question!) consider the function $g(x)=\ln x$ for $x>0$ and $g(x)=0$ for $x\le0$. Then $g$ is locally $L^1$, so defines a distribution, and $g'(x)=1/x$ for $x>0$. So $g'$ (considered as a distribution) is an admissible $u$. Therefore $$\left<u,\phi\right>=-\int_0^\infty\phi'(x)\ln x\,dx$$ works.
If you don't like derivatives, then integration by parts will give you the alternative formula $$\left<u,\phi\right>=\int_0^1\frac{\phi(x)-\phi(0)}{x}\,dx +\int_1^\infty\frac{\phi(x)}{x}\,dx.$$
Of course, this $u$ is not unique, as you can add any linear combination of the Dirac delta function (at $0$) and its derivatives to $u$.