This should be difficult in general. However there are some easy remarks to get going:

First, if $p$ is a CI point, then $X_p$ is regular. That is because when you localize, the number of generators can only drops, and it is still have to be at least $n=\dim X$. So they are equal.

Now let's try to answer number 3), when $X_0=X$? The above remark says that $X$ is non-singular. But it is more, it says that the Chow group of points is trivial.

Consider $X$ projective. Clearly the property $X_0=X$ depends on the embedding. For example, with $P^1= \text{Proj} \ k[x,y,z]/(xy-z^2)$, the point $(x,z)$ is not CI. So we just consider $X= \text{Proj}\ S/I$ with $S=k[x_0,...,x_d]$. Then a point $p$ in $\text{Proj} S$ is defined by $d$ linear forms. Modulo $I$, the number of generators drops to $n$, so $I$ must contains $d-n$ forms, and by dimension reasons, $I$ is generated by those. So $X$ must be $P^n$. (indeed, here we only needs to assume that $X_0$ is non-empty, so this also answers Question (2)).

It is harder when $X$ is affine. Except in dimension one, then we are looking for a smooth affine curve with trivial Picard group, so it must be rational curve.

I don't know much about the sCI case, or when $X_1=X$. If $X$ is smooth, we are forcing the Chow group of points to have rank one, and this should be restrictive.

As for number (1), there is a paper by Weibel, where he conjectured that for affine $X$, the set $X_0$ is a countable union of closed subsets, and solved it for dimension at most $3$.