Here is answer of (1), $c_R=2$. I assume that $R$ is complete (as $\mathfrak{m}$-adic topology) and the residue field $A/\mathfrak{m}$ is infinite (if necessary).
Let $v(R) = \dim_{R/\mathfrak{m}} \mathfrak{m}/\mathfrak{m}^2$ be the embedded dimension of $R$. Since $c_R = 2$ we have $R$ is not a regular local ring, so $v(R) \ge d+1$. Let $\mathfrak{q} = (x_1, ..., x_d)$ be a parameter ideal such that $\ell(R/\mathfrak{q}) = c_R=2$. We have $\ell(\mathfrak{m}/\mathfrak{q})= 1$, so $v(R) = d+1$. Furthermore $\mathfrak{m}^2 \subseteq \mathfrak{q}$ and $\mathfrak{m} = (x_1, ..., x_d, y)$ for some $y$.
First we consider the case $(R, \mathfrak{m})$ is unmixed i.e. $\dim R/\mathfrak{p} = d$ for all $\mathfrak{p} \in \mathrm{Ass}R$.
We note that the multiplicity $e(R) = e_R(\mathfrak{m}) \ge 2$, since $R$ is not regular.
Let $\mathfrak{q} = (x_1, ..., x_d)$ be a parameter ideal such that $\ell(R/\mathfrak{q}) = c_R=2$. It is well known that the multiplicity $e_R(\mathfrak{q}) \le \ell(R/\mathfrak{q})$. Thus we have two cases:
Case 1: $e_R(\mathfrak{q}) = 1$, so the multiplicity of $R$, $e(R) =1$. It is a contradiction.
Thus we have
Case 2: $e_R(\mathfrak{q}) = 2$, so $e(R) = 2$. Moreover $e_R(\mathfrak{q}) = \ell(R/\mathfrak{q})$, hence $(R, \mathfrak{m})$ is Cohen-Macaulay.
By Abhyankar's inequality we have $v(R) \le e(R) + d -1 = d+1$. Thus $R$ is a Cohen-Macaulay local ring of maximal embedded dimension. Furthermore, we have $\mathfrak{q}$ is a minimial reduction of $\mathfrak{m}$ and $\mathfrak{m}^2 = \mathfrak{q}\mathfrak{m}$ (by Sally?). The converse is true. Thus we have
Conclusion. Suppose $(R, \mathfrak{m})$ is unmixed. Then $c_R = 2$ if and only if $R$ is a Cohen-Macaulay (in fact, Gorenstein) of maximal embedded dimension with $v(R) = d+1$.
The case $R$ is not unmixed. Let $U$ be the largest ideal of $R$ of dimension least than $d$ (as $R$-module). Let $S = R/U$, we have $S$ in unmixed. It is clear that $c_S \le c_R = 2$. As above we have $S$ is Cohen-Macalay. So $x_1, ..., x_d$ is an regular sequence on $S$, we have $U \cap \mathfrak{q} = \mathfrak{q}U$. Therefore we have the following short exact sequence
$$0 \to U/\mathfrak{q}U \to R/\mathfrak{q} \to S/\mathfrak{q}S \to 0.$$
Since $\ell(R/\mathfrak{q}) = 2$ we have $\ell(U/\mathfrak{q}U) = \ell(S/\mathfrak{q}S) = 1$. Therefore $R$ is a regular local ring and $U = (a)$ is a principal ideal. Moreover we have $\mathfrak{m} = (x_1, ..., x_d, a)$.
Examples. Take $S = K[[X_1, ..., X_d]]$ and $T$ is a non-zero quotient ring of $S$. Let $R = S \ltimes T$ be the idealization (or trivial extension). We can check that the maximal ideal of $R$ generated by $d+1$ elements $(X_1, 0), ..., (X_d, 0), (0, 1)$. If $T = S$ we have $R$ is unmixed. If $T \neq S$ we have $R$ is not unmixed.
Edit: I give a counter-example for (2).
Let $R$ be the localization of $\mathbb{C}[X,Y]/(Y^2) = \mathbb{C}[x, y]$ at the maximal ideal $\mathfrak{m} = (x,y)$. We have $R$ is a Cohen-Macaulay local ring of dimension one, and $e(R) = 2$, $v(R) = 2$, $c_R=2$. Thus $R$ is maximal embedded dimension.
Consider the parameter ideal $\mathfrak{q} = (x^2+y)$. Since $\mathfrak{q} \nsubseteq \mathfrak{m}^2$ we have there is no parameter ideal containing $\mathfrak{q}$. Therefore $\mathfrak{q}$ is a maximal parameter ideal. One can check that
$$\ell(R/\mathfrak{q}) = \ell(\mathbb{C}[X,Y]/(X^2+Y, Y^2)) = 4.$$
