- User joshphysics has already correctly answered OP's 1st question.
2a) Concerning OP's 2nd question, one derives
$$i\hbar \delta(x-x^{\prime})~=~i\hbar\langle x | x^{\prime} \rangle ~=~\langle x | [\hat{x},\hat{p}] | x^{\prime} \rangle ~=~\langle x | \hat{x}\hat{p} | x' \rangle-\langle x | \hat{p} \hat{x} | x' \rangle$$ $$\tag{A}~=~(x-x^{\prime})\langle x | \hat{p} | x^{\prime} \rangle ~\stackrel{(1)}{=}~-i\hbar(x-x^{\prime})\frac{\partial}{\partial x}\delta(x-x^{\prime}).$$
In other words,
$$\tag{B}\delta(x-x^{\prime})~=~-(x-x^{\prime})\frac{\partial}{\partial x}\delta(x-x^{\prime}),$$
which also follows by differentiating the identity
$$\tag{C} (x-x^{\prime})\delta(x-x^{\prime})~=~0 $$
wrt. $x$.
2b) Eq. (B) should not be divide on both sides wrt. $x-x^{\prime}$. The problem is essentially that the distribution $\frac{1}{x}\delta(x)$ is ill-defined.
One argument why this is so goes roughly as follows. Recall that one way to make sense of a distribution $u$ is to evaluate on smooth test functions $g:\mathbb{R}\to \mathbb{C}$. For instance, if the distribution $u$ is the Dirac delta distribution, then by definition
$$\tag{D} u[g] ~:= ~g(0), $$
or equivalently, in a perhaps more familiar notation,
$$\tag{E} \int_{\mathbb{R}}\! dx~\delta(x) g(x) ~:= ~g(0). $$
One can in general not multiply$^1$ two distributions, but one can multiply a smooth function $f:\mathbb{R}\to \mathbb{C}$ with a distribution $u$. The product $f\cdot u$ is by definition
$$\tag{F} (f\cdot u)[g] ~:= ~u[fg]. $$
So if $u$ is the Dirac delta distribution, one gets
$$\tag{G} (f\cdot u)[g] ~:= ~f(0) g(0). $$
In OP's case, if we try to set $f(x)=\frac{1}{x}$, then $f(0)$ would be ill-defined.
Another less formal argument is that if we wrongly accept $\frac{1}{x}\delta(x)$ as a distribution, then we are prone to seemingly meaningless contradictions a la
$$x\cdot (\frac{1}{x} \delta(x))~=~x\cdot (\frac{1}{x}\cdot \delta(x))~=~(x\cdot \frac{1}{x})\cdot \delta(x)$$ $$\tag{H}~=~( \frac{1}{x}\cdot x)\cdot \delta(x)~=~ \frac{1}{x}\cdot (x\cdot \delta(x))~=~ \frac{1}{x}\cdot 0~=~0, \quad \text{(Wrong!)} $$
i.e. we have lost associativity of multiplication.
--
$^1$ We ignore Colombeau theory. See also this mathoverflow post.