Answering Question 2, I'll show that there are infinitely many $14$-dimensional commutative Frobenius algebras over an infinite field $k$ so that no two are isomorphic as algebras. Probably $14$ is not minimal.

It was shown by Suprunenko that there are infinitely many pairwise nonisomorphic $7$-dimensional commutative $k$-algebras, and in fact from the proof of this fact by Poonen in

*Poonen, Bjorn*, Isomorphism types of commutative algebras of finite rank over an algebraic closed field, Lauter, Kristin E. (ed.) et al., Computational arithmetic geometry. AMS special session, San Francisco, CA, USA, April 29–30, 2006. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4320-8/pbk). Contemporary Mathematics 463, 111-120 (2008). ZBL1155.13015. (Slightly updated version available online here.)

it follows that, among the following class of algebras parametrized by $\lambda\in k$, there are only finitely many in each isomorphism class. I'll assume that $\text{char}~k\neq2$, but an easy adjustment deals with the case $\text{char}~k=2$.

$B_\lambda$ will be the algebra with basis $\{1,x_1,x_2,x_3,x_4,s,t\}$, where $x_1^2=s$, $x_2^2=s+t$, $x_3^2=s+2t$ and $x_4^2=s+\lambda t$, and all other products of the last six basis elements are zero. The trivial extension algebras $TB_\lambda = B_\lambda\ltimes DB_\lambda$ are commutative Frobenius algebras, and I'll show, by a rather *ad hoc* argument, that $TB_\lambda\cong TB_\mu$ only if $B_\lambda\cong B_\mu$. I'm using $D$ here for the vector space dual.

Wakamatsu proved in

*Wakamatsu, Takayoshi*, **Note on trivial extensions of Artin algebras**, Commun. Algebra 12, 33-41 (1984). ZBL0537.16008.

that two finite dimensional algebras have isomorphic trivial extension algebras if and only if they are of the form $A\ltimes M$ and $A\ltimes DM$ for some algebra $A$ and $A$-bimodule $M$. Since our algebras are commutative, for us $M$ will just be an $A$-module (or a bimodule with the same action on both sides).

First, note that $\text{rad}^2B_\lambda=\text{soc}B_\lambda$ is two dimensional, spanned by $s$ and $t$.

If $TB_\lambda\cong TB_\mu$ but $B_\lambda\not\cong B_\mu$ then there must be a direct sum decomposition of vector spaces $B_\lambda=A\oplus M$ such that $A$ is a subalgebra, $M$ is a square zero ideal, and $B_\mu\cong A\ltimes DM$. Note that $M\not\cong DM$ or else $B_\lambda\cong B_\mu$. Also, $\text{soc}(A\ltimes DM)$ must be equal to $\text{rad}^2(A\ltimes DM)$ and be two-dimensional. I'll show that these things cannot happen.

Suppose they do.

If $M=S\oplus M'$ has a simple summand $S$, we can replace $A$ by $A\oplus S$ and $M$ by $M'$, so we can assume that $M$, which has radical length two, satisfies $\text{soc}M=\text{rad}M$. Since $\text{soc}(A\ltimes DM)=\text{rad}^2A\oplus\text{soc}DM$ is two dimensional, we must have $\dim\text{rad}M=\dim(M/\text{rad}M)$. It is easy to check that any two dimensional $B_\lambda$-module is self-dual, so $\dim\text{rad}M=\dim(M/\text{rad}M)=2$, and $A$ is three dimensional and $A\cap\text{soc}B_\lambda=0$.

So $\text{soc}A$ is (without loss of generality, as we can freely add multiples of $s$ and $t$ to a basis of $\text{soc}A$) a two dimensional subspace of the span of $\{x_1,x_2,x_3,x_4\}$ with square zero.

But there is no such subspace. If $x=ax_1+bx_2+cx_3+dx_4$ with $x^2=0$, then $a$ and $b$ are each determined up to two possibilities by $c$ and $d$. So in a two dimensional vector space of such elements, all pairs of values of $c$ and $d$ would have to occur, and it is easy to check that no choices of $a$ and $b$ would then give a subspace.