Does $\det(A + B) = \det(A) + \det(B)$ hold?
This does not hold true in general. For even $n$, let $A=-B$ and $\det(A) > 0$, so $\det(A+B)=0 < \det(A)+\det(B)$. Now, consider $A=B$. We have $\det(A+B)=\det(2A)=2^n \det(A) > 2 \det(A)=\det(A)+\det(B)$ for $n>1$ and $\det(A)>0$. Thus, either inequality can hold.
In general, you can't expect a formula for $\det(A+B)$. But sometimes, when you're lucky, you can use the Matrix Determinant Lemma, which says the following:
$$\det(A+uv^T)=(1+v^TA^{-1}u)\det(A),$$
where $A$ is an invertible matrix and $v^TA^{-1}u$ is interpreted as a scalar. Therefore, if $A$ is invertible, and you can write $B$ as $uv^T$ for two vectors $u,v$, then now you have a formula. One could also note that $\det(uv^T)$ is always 0.
If you're looking for a matrix operation which is well-behaved with respect to matrix addition, look for the trace.
No, there is no such law. Playing with simple matrices gives counterexamples, e.g.
$$\det\left(\begin{bmatrix} 1 & 0\\0&0\end{bmatrix}+\begin{bmatrix} 0 & 0\\0&1\end{bmatrix}\right)=\det\begin{bmatrix} 1 & 0\\0&1\end{bmatrix}=1,$$ but $$\det\begin{bmatrix} 1 & 0\\0&0\end{bmatrix}+ \det\begin{bmatrix} 0 & 0\\0&1\end{bmatrix}=0+0=0.$$