Lie-Cartan coordinates of the first kind vs Lie-Cartan coordinates of the second kind Lie-Cartan coordinates of the first kind: 
$$
  R_1 = \exp(\alpha_1w_1 + \alpha_2w_2+\alpha_3w_3)
$$
Lie-Cartan coordinates of the second kind:
$$
  R_2 = \exp(\beta_1w_1) \exp(\beta_2w_2) \exp(\beta_3w_3)
$$
My problem is are these two different?
For $\exp$ we have
$$
 b^{a+b} = b^ab^b,
$$
does this not apply to matrices?
 A: In the context of Lie groups and Lie algebras, the $w_1$, $w_2$, and $w_3$ quantities you have as arguments to the exponential are usually skew-symmetric matrices, each of which when exponentiated give rise to a rotation action in $SO(3)$. Intuitively, rotations themselves do not commute, and this gives us a good hint as to why your usage of the exponential identity is invalid.
If we expand $\exp{a}\exp{b}$, we get the product of two Taylor series:
$$\left(1 + a + \frac{a^2}{2} + \cdots \right)\left(1 + b + \frac{b^2}{2} + \cdots \right)$$
For comparison, the expansion of $\exp({a + b})$ is:
$$1 + a + b + \frac{(a + b)^2}{2} + \cdots$$
Let's look closely at the expansion of $(a + b)^2$. The result is not in general $a^2 + 2ab + b^2$ as we'll see.
$$
\begin{aligned}
(a + b)^2 &= (a + b) (a + b) \\
&= a(a + b) + b(a + b)\\
&= a^2 + ab + ba + b^2
\end{aligned}
$$
If $a$ and $b$ do not commute, the expansion of $\exp{(a + b)}$ will have the terms $\frac{ab}{2}$ and $\frac{ba}{2}$, while the expansion of $\exp{a}\exp{b}$ will have the term $ab$. In the case of $\mathfrak{so}(3)$, the group of skew-symmetric matrices you've denoted as $w_1$, $w_2$, and $w_3$, elements in this group do not commute, so $\exp{a}\exp{b} \neq \exp{(a + b)}$.
