= Solution
Separate the defining <matrix exponential> into even and odd powers and use $\tau_j^2=I_2$:
$$
e^{i\alpha_j\tau_j}=I_2\sum_{m=0}^\infty\frac{(-1)^m\alpha_j^{2m}}{(2m)!}
+i\tau_j\sum_{m=0}^\infty\frac{(-1)^m\alpha_j^{2m+1}}{(2m+1)!}
=I_2\cos\alpha_j+i\tau_j\sin\alpha_j.
$$
Therefore
$$
\boxed{A_1=\begin{pmatrix}\cos\alpha_1&i\sin\alpha_1\\i\sin\alpha_1&\cos\alpha_1\end{pmatrix},\qquad
A_2=\begin{pmatrix}\cos\alpha_2&\sin\alpha_2\\-\sin\alpha_2&\cos\alpha_2\end{pmatrix},}
$$
$$
\boxed{A_3=\begin{pmatrix}e^{i\alpha_3}&0\\0&e^{-i\alpha_3}\end{pmatrix}.}
$$
Since each $\tau_j$ is <Hermitian>, $A_j^\dagger=\cos\alpha_j I_2-i\sin\alpha_j\tau_j=A_j^{-1}$; each is a <unitary matrix>.
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