$A, B$ are linear map and dim$null(A) = 3$, dim$null(B) = 5$ what about dim$null(AB)$ $A, B$ are linear map from $\mathbb{R}^{12} \to \mathbb{R}^{12}$ and dim$null(A) = 3$, dim$null(B) = 5$ what value could dim$null(AB)$ be?
I think it could be greater than or equal to 5 because
$$
 Ker(B) \subset \{x\in \mathbb{R}^{12} : Bx \in Ker A\} = Ker(AB)
$$
but is there are better result we can get?

$\ker(AB)$ can have any dimension between $5$ and $8$.  As you have already observed, $\ker(B) \subset \ker(AB)$, so that $\dim \ker(AB) \geq 5$.  On the other hand: using the rank nullity theorem, we have
$$
\dim\operatorname{im}(AB) = 
\dim\operatorname{im}(A|_{\operatorname{im}(B)}) = 
\dim\operatorname{im}(B) - \dim \ker (A|_{\operatorname{im}(B)})\\ 
= \dim\operatorname{im}(B) - \dim [\ker (A) \cap \operatorname{im}(B)]
\\\geq \dim\operatorname{im}(B) - \dim \ker(A) = 12 - \dim \ker(B) - \dim \ker (A)
$$
So, we have
$$
12 - \dim \ker(AB) \geq 12 - (\dim \ker (B) + \dim \ker (A)) \implies\\
\dim \ker(AB) \leq \dim \ker(B) + \dim \ker(A)
$$
