Aloha :)
a) Die Divergenz eines Vektorfeldes verschwindet:$$\vec\nabla\cdot(\vec\nabla\times\vec A)=\sum_{i=1}^3\partial_i(\vec\nabla\times\vec A)_i=\sum_{i=1}^3\partial_i\left(\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}\partial_jA_k\right)$$$$\quad=\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}\partial_i\left(\partial_jA_k\right)=\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}\partial_i\partial_jA_k$$$$\quad=\partial_1\partial_2A_3+\partial_2\partial_3A_1+\partial_3\partial_1A_2-\partial_3\partial_2A_1-\partial_2\partial_1A_3-\partial_1\partial_3A_2$$$$\quad=\partial_2\partial_3A_1-\partial_3\partial_2A_1+\partial_3\partial_1A_2-\partial_1\partial_3A_2+\partial_1\partial_2A_3-\partial_2\partial_1A_3$$Wegen \(\vec A\in C^2\) gilt nach dem Satz von Schwarz: \(\partial_i\partial_jA_k=\partial_j\partial_iA_k\), sodass:$$\quad=\partial_2\partial_3A_1-\partial_2\partial_3A_1+\partial_3\partial_1A_2-\partial_3\partial_1A_2+\partial_1\partial_2A_3-\partial_1\partial_2A_3=0\quad\checkmark$$
b) Produktregel für Rotation:
$$\left(\vec\nabla\times\left(\varphi\vec A\right)\right)_i=\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}\partial_j(\varphi A)_k=\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}\partial_j(\varphi A_k)$$$$\quad=\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}((\partial_j\varphi)A_k+\varphi\partial_jA_k))=\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}((\partial_j\varphi)A_k)$$$$\quad+\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}(\varphi\partial_jA_k)$$$$\quad=\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}(\vec\nabla\varphi)_jA_k+\varphi\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}\partial_jA_k$$$$\quad=((\vec\nabla\varphi)\times\vec A)_i+\varphi(\vec\nabla\times A)_i\quad\checkmark$$
