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Sorry Alireza. I think you wanted $G(d,p)$ to be $(d+1)\times(d+1)$ upper triangular matrices over ${\mathbb F}_p$ (not $d\times d$). Then $G(d,p)$ is a witness: $d$-generated, has exponent $p$ if $p> d$, and class $d$. Thus $c(d,p)\geq d$ for $p>d$. This answers Questions 1 and 2.
Thanks Alireza. A lower bound can be established with a `witness': if $G$ is $d$-generated, has exponent $p$, and class $c$, then $c(d,p)\geq c$. For example $c(2,p)\geq 2$ if $p>2$; take $G$ to be extraspecial of exponent $p$. Your group $G(d,p)$ does not have exponent $p$ if $d>3$, and its class is $d-1$ not $d$. For Question 2 it suffices to find witnesses $G_{d,p}$ for $d=3,4$ and all $p\geq5$.
The $abc$-conjecture, and the more general $abcd$-conjecture which is needed to prove Conjectures 1 and 2 above, may be much harder to prove than my two conjectures. If they are equally hard, it would surprise me. That said, I am no expert on the $n$-term conjecture.
Thanks Jan-Christoph. Prop 2.2 of the preprint states the formula for $\sigma_p(m,n)$, but not the formula for $\pi_p(m,n)$. (If it did then the proof of $s_p(mn)\le s_p(m)s_p(n)$ would be even shorter.)
In addition to the Mathieu groups mentioned by David, there are the examples $PSL(2,p)\le S_{p+1}$ in the case $n= p+1$; and solvable examples $\Gamma L(1,2^q)\le S_{2^q}$ where $p=2^q-1$ is a Mersenne prime.
@Myerson It is important to distinguish integers and indeterminants. Let $x$ and $y$ be integers greater than 1, and let $X$ and $Y$ be indeterminants. Consider $f(X,Y), g(X,Y)\in\mathbb{Q}(Y)[X]$. Certainly $f(X,Y)\equiv 0\pmod {g(X,Y)}$ implies $f(x,y)\equiv 0\pmod {g(x,y)}$ for all integers $x,y>1$, but the converse is false. For example, $(X^2+2)\pmod {XY}$ is nonzero, but $(2^2+2)\pmod {2\cdot3}$ is zero. For this reason, I do not see that Gerry has solved the problem.
