I already know that for Borel-$\sigma$-algebras it holds that $\mathfrak B^{p+q}=\mathfrak B^p \otimes\mathfrak B^q$. Now I want to show that this is not the case for Lebesgue-$\sigma$-algebras $\mathfrak L$.
So first of all, given a zero Lebesgue-null-set $N$ in $\mathbb R^p$, I have proven that $N\times B$ is a Lebesgue-null-set in $\mathbb R^{p+q}$ for arbitrary $B\subseteq \mathbb R^q$. Now, how can I show that $$\mathfrak B^p \otimes\mathfrak B^q\subsetneq \mathfrak L^p \otimes\mathfrak L^q \subsetneq \mathfrak L^{p+q} $$
My work for the first part of the proof: So if we chose $B$ where $\mu(B)<\infty$, it is obvious that we obtain $$\lambda(N\times B)=\nu(N)\mu(B)=0\times M=0$$ where $M\in\mathbb R$ is an upper bound of $\mu(B)$.
Thus I considered the case $\mu(B)≥\infty$. Since $\mu$ is $\sigma$ -additive, we can find a sequence $B_n\subseteq B$ such that $\bigcup_{n=1}^\infty B_n = B$ and $\mu(B_n)<\infty$ for all $n$. It follows that $$\lambda(N\times B)= \lambda(N\times \bigcup_{n=1}^\infty B_n )=\nu(N)\mu(\bigcup_{n=1}^\infty B_n) = \nu(N)\lim_{n\to\infty}\mu( B_n)=0\times S=0$$ where $S$ is the upper bound of $B_n$.
I believe your proof solves completely the first part of the question. So I will only adress the second question:
$$\mathfrak L^p \otimes\mathfrak L^q \subsetneq \mathfrak L^{p+q}$$
The counterexample can be given for $p=q=1$. Then it can be extended for any indices.
Let $V$ be the Vitali set in $\mathbb{R}.$ We get that $N := V\times \{0\} \subset \mathbb{R} \times \{0\} \subset \mathbb{R}^2$ is contained in $\mathfrak L^{p+q},$ since it´s contained in a nullset.
But it´s not contained in the product Lebesgue sigma-algebra. This is because sections of sets in a product sigma-algebra must me measurable. But the section in $0$ of our set is not Lebesgue measurable, since $N_0 = V$ is the Vitali set.