Product rule of Curl 
I know how to do a to d. 
For e, i let $\phi = \frac{1}{|r|^3}$ and $A = a \times r$, tried to simplify but did not reach the answer. Anything can help. THank you
 A: rewrite $d$ with brackets: $$\vec{\triangledown }\times (\varphi \vec{A}) =[(\vec{\triangledown }\varphi)\times\vec{A})]+\varphi[\vec{\triangledown }\times\vec{A}]$$
Let's work with first addend: $[(\vec{\triangledown }\varphi)\times\vec{A})]=[\tfrac{-3\vec{r}}{\left |  \vec {r}\right |^5}\times(\vec{a}\times\vec{r})]=\tfrac{-3}{\left |  \vec {r}\right |^5}[\vec {r} \times(\vec{a}\times\vec{r})]=\tfrac{-3}{\left |  \vec {r}\right |^5}(\vec{a}\cdot\left |  \vec {r}\right |^2-\vec{r}(\vec{a}\cdot\vec{r}))$ ${\color{Red}(\color{Red}1\color{Red})}$
Remark: we use the rule: $[\vec {a} \times(\vec{b}\times\vec{c})]=\vec{b}\cdot(\vec{a}\cdot\vec{c})-\vec{c}(\vec{a}\cdot\vec{b})$
For the second addend using $(b)$ statement we get: $\varphi[\vec{\triangledown }\times\vec{A}]=\tfrac{1}{\left |  \vec {r}\right |^3}\cdot2\vec{a}$ 
       $\;\;\;\;\;\;\;\;\;\;$${\color{Red}(\color{Red}2\color{Red})}$
Now, we add $(1)+(2)$ and get your result $$\vec{\triangledown }\times (\varphi \vec{A}) = \tfrac{3}{\left |  \vec {r}\right |^5}\vec{r}(\vec{a}\cdot\vec{r})-\tfrac{1}{\left |  \vec {r}\right |^3}\cdot\vec{a}$$
