Solution:

Let $P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}$ be a polynomial with integer coefficients.

Part (i) We may write
$$P(x)=a_{0}+a_{2} x^{2}+a_{4} x^{4}+\cdots+x\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\cdots\right)$$
Define
$$Q(x)=a_{0}+a_{2} x^{2}+a_{4} x^{4}+\cdots-x\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\cdots\right)$$
Then $Q(x)$ is also a polynomial with integer coefficients and
$$P(x) Q(x)=\left(a_{0}+a_{2} x^{2}+a_{4} x^{4}+\cdots\right)^{2}-x^{2}\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\cdots\right)^{2}$$
is a polynomial in $x^{2}$.

Part (ii) We write again
$$P(x)=A(x)+x B(x)+x^{2} C(x)$$
where
$$\begin{aligned} & A(x)=a_{0}+a_{3} x^{3}+a_{6} x^{6}+\cdots \\ & B(x)=a_{1}+a_{4} x^{3}+a_{7} x^{6}+\cdots \\ & C(x)=a_{2}+a_{5} x^{3}+a_{8} x^{6}+\cdots \end{aligned}$$
Note that $A(x), B(x)$ and $C(x)$ are polynomials with integer coefficients and each of these is a polynomial in $x^{3}$. We may introduce
$$\begin{aligned} & S(x)=A(x)+\omega x B(x)+\omega^{2} x^{2} C(x) \\ & T(x)=A(x)+\omega^{2} x B(x)+\omega x^{2} C(x) \end{aligned}$$
where $\omega$ is an imaginary cube-root of unity. Then
$$\begin{aligned} S(x) T(x)=(A(x))^{2}+x^{2}(B(x))^{2}+x^{4}(C(x))^{2} & \\ & -x A(x) B(x)-x^{3} B(x) C(x)-x^{2} C(x) A(x) \end{aligned}$$
since $\omega^{3}=1$ and $\omega+\omega^{2}=-1$. Taking $R(x)=S(x) T(x)$, we obtain
$$P(x) R(x)=(A(x))^{3}+x^{3}(B(x))^{3}+x^{6}(C(x))^{3}-3 x^{3} A(x) B(x) C(x)$$
which is a polynomial in $x^{3}$. This follows from the identity
$$(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=a^{3}+b^{3}+c^{3}-3 a b c$$
Alternately, $R(x)$ may be directly defined by
$$\begin{aligned} & R(x)=(A(x))^{2}+x^{2}(B(x))^{2}+x^{4}(C(x))^{2} \\ &-x A(x) B(x)-x^{3} B(x) C(x)-x^{2} C(x) A(x) \end{aligned}$$