So I don't think I am going to answer your question here, and I think you probably know most of what I am going to write at this point. But perhaps this information will be of use to others that happen to pass by this thread.

Let $G$ be algebraic (including compact, and complex reductive) and denote $R(G)=\mathrm{Hom}(\pi,G)$.

The tangent space of $R(G)$ at $\rho$ always make sense since it is algebraic and it is the twisted cocycles $Z^1(\pi,\mathfrak{g}_{\mathrm{Ad}\rho})$.

Likewise, the tangent space to a conjugation orbit also generally makes sense and is the
twisted coboundaries $B^1(\pi,\mathfrak{g}_{\mathrm{Ad}\rho})$.

Both of these are results of Weil from *Remarks on cohomology of groups*, I believe.

Unfortunately the tangent space to a quotient is not always the quotient of the tangent spaces, so it is not true that the tangent space to $R(G)/G$ is always $H^1(\pi,\mathfrak{g}_{\mathrm{Ad}\rho})$. Note that if $G$ is compact this quotient is the orbit space, and if $G$ is complex reductive it is the GIT quotient.

An easy example is the following: Let $\pi=\mathbb{Z}$ and let $G=\mathrm{SL}(n,\mathbb{C})$. Then the GIT quotient $R(G)/G$ is $\mathbb{C}^{n-1}$, parameterized by coefficients of the characteristic polynomial, and so is smooth. Thus the tangent space to the identity character is $\mathbb{C}^{n-1}$. On the other hand, the orbit of the identity is trivial and it is a smooth point in $R(G)=G$. So the cocycles are $\mathfrak{g}=\mathbb{C}^{n^2-1}$ and the coboundaries are trivial. Thus the first cohomology is also $\mathbb{C}^{n^2-1}$ which is much bigger than the correct result of $\mathbb{C}^{n-1}$.

However the following is true. If $\rho \in R(G)$ is smooth (always true for irreducible reps in surface groups, and all reps in twisted surface groups and all reps in free groups), and also have a closed conjugation orbit, then using an appropriate Slice Theorem (Luna, or Mostow) one can prove that the tangent space at the equivalence class $[\rho]$ is the tangent space at 0 in a quotient of cohomology: $T_0(H^1(\pi,\mathfrak{g}_{\mathrm{Ad}\rho})/C)$, where $C$ is the centralizer of $\rho$ in $G$. But in general there are known counter examples. See here.

Returning to the above example, we then have $\mathfrak{sl}(n,\mathbb{C})/C=\mathbb{C}^{n-1}$ since $C=\mathrm{SL}(n,\mathbb{C})$ (still parameterized by the coefficients of the characteristic polynomial, but now instead of the determinant being fixed, the trace is fixed).

If the centralizer is acts trivially one recovers the usual statement that the tangent space is $H^1(\pi,\mathfrak{g}_{\mathrm{Ad}\rho})$. For example, this holds for irreducible representations and $G=\mathrm{GL}(n,\mathbb{C})$ or $\mathrm{SL}(n,\mathbb{C})$ (or their maximal compact subgroups), and $\pi$ is a free group or a (twisted) surface group. But even when $\pi$ is a free group and $G=\mathrm{PSL}(2,\mathbb{C})$ and $\rho$ is an irreducible representation there are counter examples (namely there are irreducible singularities in that case).

There is always an open dense subset of smooth irreducibles and a further subset of those whose centralizer is equal to the center of $G$. These are called "good" representations (a la Millson). The set of good representations modulo G is a manifold, and the set of all irreducibles forms an orbifold (for general $G$ irreducible means that $\rho(\pi)$ is not contained in a parabolic subgroup).

Let me try to briefly address (1). Whenever the dimension of the tangent space is the dimension of the space, in the strong topology there is a neighborhood that looks like a ball. The converse is not true in general.

Think of $x^2=y^3$ which has a singularity at $(0,0)$ algebraically but the neighborhood of $(0,0)$ in the variety is homeomorphic to a Euclicean neighborhood. On the other hand $xy=0$ also has a dimension jump in tangent space at $(0,0)$ but no neighborhood is homeomorphic to a ball. Anyway, singularities in an algebraic setting can be mild or wild or something in between (like orbifold type).

With respect to (1), if $[\rho]$ is in the interior of a maximal simplex, I think it might still depend on $[\rho]$. Anyway, in general (as the example $x^2=y^3$ shows) knowing that a point in a semi-algebraic set has a neighborhood homeomorphic to a ball does not tell you it is not singular.

Note: A recent paper of Millson and Kapovich shows that singularities in character varieties can get as bad as you can imagine. See here.

For $\pi$ a free group however, I believe generally (for all but a finite number of counter examples) the situation is this: reducibles are singular and have no neighborhood homeomorphic to a ball. For $\pi$ a free group and $G$ equal to $\mathrm{GL}(n,\mathbb{C})$ or $\mathrm{SL}(n,\mathbb{C})$ this more or less has been established. See my paper here. Also the irreducibles are either smooth or admit orbifold singularieties. But there are orbifolds that are homeomorphic to manifolds (but not always), so it seems that one could have an orbifold singularity (with an irreducible) that happens to have a neighborhood homeomorphic to a ball (this would give a negative answer to (1)). I don't however have an example to show this for certain.

About (2), if the cohomology corresponds to the tangent space (like for free groups, or twisted surface groups), then yes, at a singularity, the dimension must jump, and hence the cohomology does too. But if the point does not correspond to cohomology to begin with, then although the tangent space must still have a dimension jump, the cohomology conceivably could not. Again, I don't have an example to show this for certain.