= Solution
The fixed points of reciprocal root pairing are $1$ and $-1$, so part (iii) alone does not establish their parity. Use the polynomial identity itself.
Write $P(x)=(x-1)^rA(x)$, where $A(1)\ne0$ and allow $r=0$ if one is not a root. After substituting in $P(x)=x^{2m}P(1/x)$ and canceling $(x-1)^r$ away from $x=1$, obtain
$$
A(x)=(-1)^r x^{2m-r}A(1/x).
$$
Letting $x\to1$ yields $A(1)=(-1)^rA(1)$, so $r$ is even.
Similarly write $P(x)=(x+1)^sB(x)$ with $B(-1)\ne0$. This time cancellation gives $B(x)=x^{2m-s}B(1/x)$. Letting $x\to-1$ yields $B(-1)=(-1)^{2m-s}B(-1)$, so $s$ is even. Thus
$$
\boxed{\text{Both }1\text{ and }-1\text{ have even algebraic multiplicity.}}
$$
The conclusion includes multiplicity zero. It is a consequence of the reciprocal identity in even degree, which is stronger at these fixed points than just pairing the roots.
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