Does "finitely presented" mean "always finitely presented", considered in general

It is true for any algebras (sets with operations) satisfying any set of identities (laws). Indeed, if $X, Y$ are finite generating sets $R=\{u_i,v_i\mid i\in I\}$ be a finite set of defining relations over $X$, $Q$ is any set of defining relations over $Y$, then for every relation from $R$ there exists a proof of that relation using relations from $Q$. The proof involves a finite number of elements of $Q$. If $Q'$ is the (finite) collection of all elements of $Q$ involved, then the algebra has a finite presentation $\langle Y, Q'\rangle$.

Yes in general.

See Adámek and Rosicky, Locally Presentable and Accessible Categories Cambridge University Press, Cambridge, (1994).

T. 3.12 p. 143.

Of course "in general" I mean: every "algebraic theory" (many sorted) Set models.

For topological algebraic structures, (like profinite groups) this equivalence isn't true for the Yiftach Barnea example

This is not a definite answer, but I suspect that for profinite groups this is wrong (here generating means in the topological sense: generates a dense subgroup). The reason is that there are finitely generated profinite groups which have an infinite minimal generating subset. For example if you look at $\Pi_{n \geq 5}A_n$, then it is $2$-generated as a profinite group, but it also has a minimal infinite generating subset. The point is that profinite words could be infinite and may contain infinite number of distinct letters.

It would be nice to see an example.

EDIT: First, a simpler example of a profinite group which is cyclic, but has a infinite minimal generating subset is $\hat{\mathbb{Z}}$. Now, look at the category of abelian profinite groups. We can look at $\hat{\mathbb{Z}}$ as a quotient of $\hat{\mathbb{Z}}^2$ by $\hat{\mathbb{Z}}$. Then we found a presentation of $\hat{\mathbb{Z}}$ with two generators and in one case with one relation and in the other case with infinite number of relations.

Admittedly, this is not the minimal number of generators. Maybe this can be improved.

Final Answer (I hope): Okay, so I can even give an example with one generator. Take $G=\Pi_p \mathbb{Z}/ p \mathbb{Z}$, where $p$ goes over all the primes, in the category of profinite groups or the category of abelian profinite groups. $G \cong \hat{\mathbb{Z}}/N$, where $N=\Pi_{p} p \mathbb{Z}_p$, where $\mathbb{Z}_p$ are the $p$-adic integers. Now, $p\mathbb{Z}_p \cong \mathbb{Z}_p$. So, $N \cong \Pi_p \mathbb{Z}_p \cong \hat{\mathbb{Z}}$ and again $N$ is one-generated but has an infinite minimal generating subset.

I cannot add a comment now for some reason, so Richard thanks a lot!