If a commutative graded algebra is free over a graded subalgebra, then must it have a graded basis? 
Fix a field $\mathbf{k}$ and an $\mathbb{N}$-graded commutative $\mathbf{k}$-algebra $A = \bigoplus\limits_{n = 0}^{\infty} A_n$ of finite type. ("Finite type" means that each $A_n$ is a finite-dimensional $\mathbf{k}$-vector space. All algebras are understood to contain $1$.) Let $B = \bigoplus\limits_{n = 0}^{\infty} B_n$ be an $\mathbb{N}$-graded $\mathbf{k}$-subalgebra of $A$ (so that $B_n \subseteq A_n$ for all $n$). Assume that $A$ is free as a $B$-module.
Question 1. Does it follow that the $B$-module $A$ has a basis consisting of homogeneous elements?
Question 2. If yes: Can we omit the requirement that $A$ is of finite type?
Question 3. If yes: Does this still hold if $\mathbf{k}$ is a commutative ring rather than a field?
Question 4. If no, does it help to assume that $A$ is a graded subalgebra of a polynomial ring over $\mathbf{k}$?
Question 5. Does it help if $A_0 \cong \mathbf{k}$ and the $\mathbf{k}$-algebra $A$ is generated by $A_1$ ?

Sorry for the onslaught of questions -- I am hoping that an answer to one will likely solve most of the others, which is why I prefer not to split them across several topics.
The question cloud originates from working with Vic Reiner on cyclic quasisymmetric functions, but I find it more fundamental. I originally thought these would be easy exercises in graded linear algebra (using projection maps, linear combinations and recurrent constructions), but I see no obvious point of vantage for such tactics.
 A: 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.
