Q1: no (this makes Q2, Q3 obsolete)
Q4, Q5: yes (for $k$ a field)
Example for Q1: Let $B = k \oplus k$ (componentwise operations) be concentrated in degree zero and take $$A = A_0 \oplus A_1 \oplus A_2 := B \oplus k \oplus k$$ $B$ acts on $A_i=k$ via the projection $p_i:B \twoheadrightarrow k$. This makes $B \cong k \oplus k$ as $B$-modules. Make $A$ into a graded ring by letting the products of positive degree be zero. As $B$-module, $A\cong B \oplus B$ is free. But it has no homogeneous base (otherwise the homogeneous components would be direct sums of $B$).
In general we can say:
Let $B$ be a non-negatively graded ring (not necessarily commutative) such that each projective $B_0$-module is free. Then each bounded below, graded $B$-module $M$ which is free as $B$-module has a homogenous base.
If $M$ is of finite type, it's enough that each finitely generated projective $B_0$-module is free.
Sketch of proof: Let $B_+ = \oplus_{n > 0}B_n$ be the irrelevant ideal. Since $M$ is free as $B$-module, $M/B_+M$ is free as $B_0$-module. It's also a graded $B_0$-module. Hence the homogeneous summands of $M/B_+M$ are direct summands of a free $B_0$-module and hence projective. By the assumptions on $B_0$ the homogeneous summands are free $B_0$-modules. Now the result follows verbatim from Manny Reyes' answer in https://math.stackexchange.com/questions/557402/graded-free-is-stronger-than-graded-and-free. I should add that the projective-free argument is due to Eric Wofsey taken from the same link. q.e.d.
In Q4, Q5 we have $A_0 = B_0 = k$ so the result applies, if $k$ is a field.
