= Solution
The supplied <matrix> obeys
$$
J_3^2=\operatorname{diag}(1,1,0),\qquad J_3^3=J_3.
$$
Separate the even and odd powers in the <matrix exponential> to obtain
$$
\boxed{e^{-i\theta J_3}=I-i\sin\theta J_3+(\cos\theta-1)J_3^2
=\begin{pmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix}.}
$$
It fixes the $z$ component and is a counterclockwise active <rotation> viewed from $+z$. This sign differs from the clockwise convention in <4/b>, because the source fixes the sign of both $J_3$ and the exponential. Replacing $\theta$ by $-\theta$ yields the clockwise <rotation>.
With $(J_k)_{ij}=-i\epsilon_{kij}$, the <generators of rotations> obey
$$
\boxed{[J_i,J_j]=i\epsilon_{ijk}J_k.}
$$
The <commutator> measures the leading difference between performing two infinitesimal <rotations> in opposite orders. For example, with $X=-i\alpha J_1$ and $Y=-i\beta J_2$,
$$
e^Xe^Ye^{-X}e^{-Y}=I+[X,Y]+O(\alpha^2\beta,\alpha\beta^2)
=I-i\alpha\beta J_3+O(\alpha^2\beta,\alpha\beta^2).
$$
The failure to commute is itself an infinitesimal <rotation> about the perpendicular axis; the <Lie algebra> closes on the same three <generators of rotations>.
To pass from this local statement to the full <group> property, put $A_k=-iJ_k$. These are real antisymmetric <matrices>, with $[A_i,A_j]=\epsilon_{ijk}A_k$. The <Baker--Campbell--Hausdorff formula> shows locally that the logarithm of a product of their exponentials remains a linear combination of the $A_k$, because all nested <commutators> do. Globally, each exponential has $R^TR=I$ and $\det R=e^{\operatorname{tr}A}=1$. If $R_1,R_2$ have these properties, then
$$
(R_1R_2)^T(R_1R_2)=R_2^TR_2=I,\qquad \det(R_1R_2)=1.
$$
Every real three-dimensional <orthogonal matrix> of <determinant> $+1$ has an eigenvector with <eigenvalue> $+1$: its nonreal <eigenvalues> come in conjugate pairs, and the product of all three is $+1$. On the perpendicular plane it acts as a two-dimensional <rotation>. It therefore has axis-angle form and can be written $e^{-i\theta\hat n^kJ_k}$. Thus the product is again a <rotation>. The identity is $\theta=0$, the inverse reverses $\theta$, and associativity comes from <matrix multiplication>. This proves that all these <rotations> form the <special orthogonal group> $SO(3)$. The <commutator> supplies local closure, while the orthogonality and axis-angle argument supplies the necessary global step.
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