In abelian groups: $$\text{soc}\left(\prod_{p\text{ prime}} \mathbb{Z}/p\mathbb{Z}\right) = \bigoplus_{p\text{ prime}} \mathbb{Z}/p\mathbb{Z}\not\cong \prod_{p\text{ prime}}\mathbb{Z}/p\mathbb{Z} = \prod_{p\text{ prime}} \text{soc}(\mathbb{Z}/p\mathbb{Z})$$
so the socle functor does not preserve limits and thus does not have a left adjoint. I also doubt that the radical functor preserves infinite products, but I don't have an example off the top of my head.

Also, we have a coequalizer diagram $$\mathbb{Z}\rightrightarrows \mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$$ where the arrows on the left are the identity and the multiplication by $4$ map. We have $\text{rad}(\mathbb{Z}) = \text{soc}(\mathbb{Z}) = \{0\}$ and $\text{rad}(\mathbb{Z}/4\mathbb{Z}) = \text{soc}(\mathbb{Z}/4\mathbb{Z}) = \{0,2\}$. So taking radicals or socles gives $$\{0\}\rightrightarrows \{0\} \to \{0,2\}$$
which is not a coequalizer diagram. So neither functor preserves colimits, and neither has a right adjoint.