$a,b \in\mathbb{Z}$ and $a$ is even, show that if $a^{2}b$ not divisible by $8$, then $b$ is odd 
The fourth question is how can I do?

Q: Given that $a, b \in \mathbb{Z}$ and $a$ is even, show that if $a^2b$ is not divisible by $8$, then $b$ is odd.

 A: If $a$ is even, then there exists $n$ such that $a=2n$. Hence, $a^2b= 4n^2b$.

Can you take it from here, knowing that $8=2^3$?
A: Since $a$ is even number, let us assume it as, $a=2n$. Hence $a^2\cdot b = (2n)^2\dot b = 4n^2\cdot b$.
Now $a^2\cdot b$ is divisible by $8$ if $a^2 \cdot b = 8k$ where $k \in \mathbb{Z}$. This is possible only if $b$ is even. Hence $a^2b$ is not divisible by $8$ if $b$ is odd.
A: Is $a$ singly even or doubly even? If the former, that means $a = 4k + 2$, while the latter means $a = 4k$. If $a$ is singly even, then $a^2$ is doubly even, since $a^2 = (4k + 2)(4k + 2) = 16k^2 + 8k + 4$; however, $a^2$ is not divisible by $8$.
If $a$ is doubly even, then so is $a^2$, and what's more, it's also divisible by $8$, in which case the parity of $b$ is irrelevant, $8$ will be a divisor of $a^2 b$.
But if $a$ is singly even and $b$ is odd, then $a^2 b$ will be divisible by $4$ but not by $8$.
A couple of concrete examples: suppose $a = 10$ and $b = 3$. Then $300$ is divisible by $4$ but not by $8$. Suppose $b = 2$ instead. Then $200$ is divisible by $4$ and by $8$.
A: HINT I would argue by contradiction as I think this is easier here: suppose both $a$ and $b$ are even ... can you show that $a^2b$ is then divisible by $8$? 
So what gives if $a^2b$ is not divisible by $8$?
