1) The transformation matrix is in general 4x4: $\begin{equation} \begin{bmatrix} R&t\\ 0&1\\ \end{bmatrix} \end{equation}$ with $R$ being a 3x3 rotation matrix and $t$ being a translation vector.
Let $x_1,x_2,x_3\in\mathbb{R}^3$ be the first triangle and $\bar{x}_1,\bar{x}_2,\bar{x}_3\in\mathbb{R}^3$ the second one.
Then following Arun et al. 1987, we get:
$\begin{equation} c = \frac{1}{3}(x_1 + x_2 + x_3), \quad \bar{c} = \frac{1}{3}(\bar{x}_1 + \bar{x}_2 + \bar{x}_3), \quad\text{(calculate centroids)} \end{equation}$
$\begin{equation} y_i = x_i-c, \quad \bar{y}_i = \bar{x}_i-\bar{c}, \quad\text{(subtract centroids from points)} \end{equation}$
$\begin{equation} H = y_1\bar{y}_1^\top + y_2\bar{y}_2^\top + y_3\bar{y}_3^\top, \quad U\cdot\Sigma\cdot V^\top = H \quad\text{(singular value decomposition)} \end{equation}$
$\begin{equation} R = V U^\top, \quad t = \bar c - Rc \end{equation}$
2) The solution above is a least-squares fit, so if there is a tiny deformation this should not affect the result too much. It is hard to say more about it, if we don't know what kind of deformation we are expecting (scale change only, affine deformation, or something else).