If $I+J=R$, where $R$ is a commutative rng, prove that $IJ=I\cap J$.

The statement you are trying to prove is only necessarily true for commutative rings with $1$. In this case, you can argue that $$I\cap J\subseteq (I\cap J)R=(I\cap J)(I+J)=I(I\cap J)+ J(I\cap J)\subseteq IJ+ IJ=IJ$$

A counterexample to the statement for general rngs is given by endowing the group $\mathbb Z$ with the zero product, that is defining $a\cdot b=0,\forall a,b\in\mathbb Z$. The ideals $(2)$ and $(3)$ are still comaximal, as for any $x\in\mathbb Z$ we can write $x=a\times 2+b\times 3$ for some $a,b\in\mathbb Z$, where $\times$ denotes regular multiplication, and $a\times 2=2+\cdots+2\in(2)$ where addition is performed $a$ times, and similarly $b\times 3\in (3)$. But $(2)\cap (3)=(6)\neq (0)$, yet clearly $(2)(3)=(0)$.