A modal algebra is an algebra $\langle M,\vee,\wedge, ',0,1,^\circ\rangle$ such that $\langle M,\vee,\wedge, ',0,1,\rangle$ is a Boolean algebra and the operation $^\circ$ satisfies $(x\wedge y)^\circ= x^\circ\wedge y^\circ$ and $1^\circ=1$.
In the paper "Pretabular Varieties of Modal Algebras" of W. J. Blok on page 110, I faced with the following claim:
Let $A$ be a $0$-generated subalgebra of a modal algebra (or Boolean algebra). Then it is the free object on $0$-generators in the variety $V(A)$.
I did the following steps to prove this claim. Let $C\in V(A)$, there is only empty map from $\emptyset$ to $C$ and $\emptyset$ to $A$. Thus, to satisfy the universal mapping property, we need only find a homomorphism $\varphi: A\to C$. Now, from the fact $C\in HSP(A)$, we have $E_1\in P(A)$ and $E_2\in S(E_1)$ for which $C\in H(E_2)$. If we could find a homomorphism from $E_1$ to $E_2$, the $\varphi$ could be easily obtained, but I think this homomorphism does not generally exist. However, it exists if we can show that $E_1$ is the only non-empty subalgebra of itself.
After proving this claim, I need to know if a similar claim is true for an $m$-generated subalgebra of an arbitrary algebra.
For a Boolean algebra $A$ generated by $m$ ($m>1$) elements the claim seems to be trivial (as it is mentioned in first answer) since $V(A)$ is equal to the class of all Boolean algebras, refer to here.
As to the last question, the answer is yes for finite m. We have:
Let A be an m-generated subalgebra of a Boolean algebra (i.e. of 2^(2^m) members). Then it is the free object on m-generators in the variety V(A).