Answer
Property $2$: $\vec {a} \cdot \vec {b}=\vec {b} \cdot \vec {a}$
Property $4$: $(c\vec {a}) \cdot \vec {b}=c (\vec {a} \cdot \vec {b})$
Property $5$: $\vec {0} \cdot \vec {a}=0$
Work Step by Step
Proof of Property $2$.
Let $a= \lt a_1,a_2,a_3 \gt$ and $b= \lt b_1,b_2,b_3 \gt$
$\vec {a} \cdot \vec {b}=a_1b_1+a_2b_2+a_3b_3$
Using commutative law:
$\vec {a} \cdot \vec {b}=a_1b_1+a_2b_2+a_3b_3=b_1a_1+b_2a_2+b_3a_3$
Thus, $\vec {a} \cdot \vec {b}=\vec {b} \cdot \vec {a}$
Proof of Property $4$.
Let $a= \lt a_1,a_2,a_3 \gt$ and $b= \lt b_1,b_2,b_3 \gt$
$(c\vec {a}) \cdot \vec {b}=ca_1b_1+ca_2b_2+ca_3b_3=c(a_1b_1+a_2b_2+a_3b_3) $
$(c\vec {a}) \cdot \vec {b}=c (\vec {a} \cdot \vec {b})$
Thus,$(c\vec {a}) \cdot \vec {b}=c (\vec {a} \cdot \vec {b})$
Proof of Property $5$.
Let $0= \lt 0,0,0 \gt$ and $a= \lt a_1,a_2,a_3 \gt$
$\vec {0} \cdot \vec {a}=\lt 0,0,0 \gt \cdot \lt a_1,a_2,a_3 \gt$
$\vec {0} \cdot \vec {a}=0a_1+0a_2+0a_3$
Thus, $\vec {0} \cdot \vec {a}=0$