Results (assuming parameter interval pmi=(T0,1) and parameter t = T)

(NumericalApproximation[Root, (RowDefault["0.777", FEExpressionsNumericalApproximationElider]), Root[-1+12 #1-20 ((#1)^2)+8 ((#1)^3)&,2], ])

T0=Root[8T^3-20T^2+12T-1,2];N[T0,4]

OO=Sqrt[(4-11 (T^1)+8 (T^2))/T];

γγ=((5-24 (T^1)+23 (T^2)+4 (T^3)-6 (T^4)+2 (T^5))/(98 (((-1+T))^3) (T^2) ))+(((1-4 (T^1)+5 (T^2)-6 (T^3)) Ω)/(98 (((-1+T))^3) (T^2) ));

ββ=((3-31 (T^1)+135 (T^2)-328 (T^3)+484 (T^4)-440 (T^5)+240 (T^6)-72 (T^7)+8 (T^8))/(8 (((-1+T))^3) (T^2)   (-1+5 (T^1)-6 (T^2)+(T^3))))+(((-1+9 (T^1)-31 (T^2)+52 (T^3)-44 (T^4)+16 (T^5)) Ω)/(8 (((-1+T))^3) (T^2)  (-1+5 (T^1)-6 (T^2)+(T^3))));

Limit[γγ/.Ω->OO,T->1,Direction->1]

RootReduce[(γγ/.Ω->OO/.T->T0)]-1

Limit[ββ/.Ω->OO,T->1,Direction->1]

RootReduce[(ββ/.Ω->OO/.T->T0)]-RootReduce[(1+2Tan[π/14]^2)^2]

Checks, now symbolically

p1=(4-11 (T^1)+8 (T^2))-T (Ω^2);
p2=(5-24 (T^1)+23 (T^2)+4 (T^3)-6 (T^4)+2 (T^5))+Ω((1-4 (T^1)+5 (T^2)-6 (T^3)))-C (98 (((-1+T))^3) (T^2) );
p3=(3-31 (T^1)+135 (T^2)-328 (T^3)+484 (T^4)-440 (T^5)+240 (T^6)-72 (T^7)+8 (T^8))+Ω(-1+9 (T^1)-31 (T^2)+52 (T^3)-44 (T^4)+16 (T^5))-B(8 (((-1+T))^3) (T^2)  (-1+5 (T^1)-6 (T^2)+(T^3)));

qv1=Collect[Factor[GroebnerBasis[{p1,p2,p3,T(T-1)(-1+5 T-6 (T^2)+(T^3))u+1},{B,C},{u,Ω,T}]][[1]],B]

qv1b=Collect[Factor[GroebnerBasis[{p1,p2,p3,T(T-1)(-1+5 T-6 (T^2)+(T^3))u+1,β^2-B,γ^2-C},{β,γ},{u,Ω,T,B,C}]][[1,1]],β]

p7[x_]=x^7+(-7s)x^6+a5 x^5+a4 x^4+a3 x^3+a2 x^2+a1 x+a0

Solve[{p7[1]+L==0,p7[-1]+L==0},{a0,a1}]

p7a[x_]=p7[x]/.{a0->-a2-a4-L+7 s,a1->-1-a3-a5}

Coefficient Comparison for obtaining polynomials (all should vanish!)

L1=Drop[CoefficientList[Numerator[Together[((1-x^2)(x-α)(x-β)/(49(x-γ)^2)D[p7a[x],x]^2)-(L^2-p7a[x]^2)]],x]/.α->(2 s-β+2γ),-1];

Tailoring the ideal by known values for ,α,β

L2={p7a[2s-β+2γ]+L,p7a[β]-L};

LL=Join[L1,L2];

Union[Cases[LL,_Symbol,∞]]

q2=Collect[Factor[GroebnerBasis[Join[L1,L2,{(γ^2-1)(49γ^2-1)(49γ^2-9)(49γ^2-25)u+1}],{β,γ},{u,a2,a3,a4,a5,s,L}]][[1]],β]

qv1b-q2

The Weierstrass normal form for the p(B,CC)=0 genus 1 elliptic planar curve (check it via Maple algcurves)

wf=-1518+(b0^2)-225 c0+3 (c0^3)=0

b0/.Solve[-1518+(b0^2)-225 c0+3 (c0^3)==0,b0][[-1]]

a symbolic  check for p(βs,γs)=p(B,CC)=0

Together[qv1/.{B->ββ,C->γγ}/.{Ω->OO}]

a symbolic  check for p(β,γ)=0

Lemma βγ=lexpr ∧ p(β,γ)=p1(B,CC)+p2(B,CC)βγ

lexpr=1/(28 T^2 (T-1)^3)((1-2 T-7 (T^2)+18 (T^3)-12 (T^4)+4 (T^5))+Ω((-1)(-1+6 T-9 (T^2)+6 (T^3))))

Together[ββ γγ-lexpr^2/.Ω->OO]

q2b=Collect[Expand[q2/.{β->Sqrt[B],γ->Sqrt[C]}/.{(B)^(3/2)->(B)S2,(B)^(5/2)->(B)^2S2,(B)^(7/2)->(B)^3S2,
(B)^(9/2)->(B)^4S2,
(B)^(11/2)->(B)^5S2,
(C)^(3/2)->(C)S1,
(C)^(5/2)->(C)^2S1,
(C)^(7/2)->(C)^3S1,
(C)^(9/2)->(C)^4S1,
(C)^(11/2)->(C)^5S1
}/.{Sqrt[B]->S2,Sqrt[C]->S1}],{S2,S1}]

Together[RootReduce[q2b/.{S1 S2->lexpr}/.{B->ββ,C->γγ}/.Ω->OO]]