Please help, I have no idea where to start:
Let $m, n \in \mathbb{N}.$ Then $m$ and $n$ are complementary if either $\mathcal{P}$ halts when executed on input $x$ or $\mathcal{Q}$ halts when executed on input $x$ -- or both -- for all $x \in \mathbb{N}$, when $\mathcal{P}$ is the $\mathcal{L}$-program such that $\#(\mathcal{P}) = m$ and $\mathcal{Q}$ is the $\mathcal{L}$-program such that $\#(\mathcal{Q}) = n.$
Let $$\text{Comp} = \{ \langle m, n \rangle \mid m,n \in \mathbb{N} \text{ and $m$ and $n$ are complementary} \} \subseteq \mathbb{N}.$$
Use a many-one reduction -- or, if possible, a 1-reduction -- to prove that the set Comp is not recursive.
First note that the set $T$ of numbers $x$ whose program halts for every input is not recursive (by Rice's Theorem). Let $e$ be a number whose program never halts, for every input. Then, $x\in T$ iff $\langle x,e\rangle\in \text{Comp}$. So, $T\leq_1\text{Comp}$.