= Solution
Start with an <elliptic curve> having a rational point $T$ of order $2$. Move that point to $(0,0)$ and take an integral model
$$
E:y^2=x(x^2+ax+b),\qquad a,b\in\mathbb Z,\quad b(a^2-4b)\ne0.
$$
Put $b'=a^2-4b$ and
$$
E':Y^2=X(X^2-2aX+b').
$$
The <two-isogeny formula> and its dual are
$$
\phi(x,y)=\left(x+a+\frac bx,\ y\left(1-\frac b{x^2}\right)\right),
$$
$$
\widehat\phi(X,Y)=\left(\frac{X-2a+b'/X}{4},\ \frac{Y(1-b'/X^2)}8\right).
$$
They extend to the projective curves, with kernels $\{O,T\}$ and $\{O,T'\}$, where $T'=(0,0)\in E'$. Substitution verifies their target equations, and the <elliptic-curve addition formula> verifies $\widehat\phi\phi=[2]$ and $\phi\widehat\phi=[2]$. The division by $4,8$ in the dual identifies the twice-transformed curve, with coefficients $4a,16b$, with $E$.
Define the <two-torsion square-class homomorphisms>
$$
\alpha(O)=1,\quad\alpha(T)=[b],\quad\alpha(x,y)=[x],
$$
$$
\alpha'(O)=1,\quad\alpha'(T')=[b'],\quad\alpha'(X,Y)=[X].
$$
Question 1(ii) proves these are homomorphisms, since $f'(0)=b$. Their kernels are
$$
\boxed{\ker\alpha=\widehat\phi(E'(\mathbb Q)),\qquad\ker\alpha'=\phi(E(\mathbb Q)).}
$$
For completeness, the first coordinate of the dual is $Y^2/(4X^2)$, so a nonexceptional dual image has square $x$-coordinate. Conversely, if $x=u^2\ne0$, the preimage equation is
$$
X^2-(4x+2a)X+b'=0,
$$
whose discriminant is $16(x^2+ax+b)=16(y/u)^2$. The roots $X=2x+a\pm2y/u$ are rational and nonzero. Choosing $Y=\pm2uX$ gives a point on $E'$, and a choice of sign makes its dual image exactly $(x,y)$. The point $T$ has a rational dual preimage precisely when $b$ is a square, as seen from the roots $a\pm2\sqrt b$ of the nonzero two-torsion polynomial on $E'$. The identity is already a dual image. Applying this argument to the transformed curve proves the second kernel assertion as well, because scaling an $x$-coordinate by $4$ does not change its <square class>.
The <two-isogeny descent> is finite because an image square class has a signed <square-free integer> representative dividing $b$. Indeed, if $\ell\nmid b$ and $v_\ell(x)>0$, the other factor $x^2+ax+b$ is a unit, so $v_\ell(y^2)=v_\ell(x)$ forces an even valuation. If $v_\ell(x)<0$, the term $x^2$ dominates that factor and $v_\ell(y^2)=3v_\ell(x)$, again forcing even valuation. Thus odd valuations can occur only at primes dividing $b$. The same argument applies to $b'$ on $E'$.
For each candidate signed squarefree divisor $d\mid b$, put $x=d(U/V)^2$, with coprime integers $U,V$. The curve equation becomes the <two-isogeny descent quartic>
$$
\boxed{N^2=dU^4+aU^2V^2+(b/d)V^4.}
$$
A solution with $UV\ne0$ yields $x=dU^2/V^2$ and $y=dUN/V^3$. Conversely, every point in that class gives such a primitive integer solution, since a rational square root of an integer is integral. The boundary solutions $U=0$ and $V=0$ account respectively for the classes of $T$ and $O$. Rational solutions prove that a class occurs; a real or congruence obstruction excludes it. Merely finding local solutions everywhere does not automatically prove a rational solution.
To extract the rank, write $A=\alpha(E(\mathbb Q))$ and $A'=\alpha'(E'(\mathbb Q))$. The <Mordell-Weil theorem> gives
$$
|E(\mathbb Q)/2E(\mathbb Q)|=2^r|E(\mathbb Q)[2]|.
$$
The isogeny factorization gives the index $|A|$ for the first quotient by $\widehat\phi E'$. For the remaining index, apply $\widehat\phi$ to $E'/\phi E$. Its kernel has size $2/e$, where
$$
e=|\ker\widehat\phi\cap\phi E(\mathbb Q)|=|\phi(E(\mathbb Q)[2])|=|E(\mathbb Q)[2]|/2.
$$
The equality follows because $\widehat\phi(\phi P)=2P$. Consequently
$$
|E/2E|=|A|\,|A'|\,e/2=|A|\,|A'|\,|E[2]|/4,
$$
and cancellation yields the <two-isogeny rank formula>
$$
\boxed{2^r=\frac{|A|\,|A'|}{4}.}
$$
This formula is valid whether there is just one rational nonzero two-torsion point or all three.
For the first curve, $a=-1,b=1$, and the isogenous curve is
$$
E':Y^2=X(X^2+2X-3).
$$
The only candidate square classes on $E$ are $1,-1$. Since $x^2-x+1>0$ for every real $x$, a real affine point has $x\geq0$; the exceptional torsion class is $[b]=1$. Thus
$$
A=\{1\}.
$$
On $E'$, the candidates are $1,-1,3,-3$. They all occur: the identity gives $1$, $T'$ gives $-3$, $(3,6)$ gives $3$, and $(-1,2)$ gives $-1$. Hence $|A'|=4$, and
$$
\boxed{\operatorname{rank}\bigl(y^2=x(x^2-x+1)\bigr)=0.}
$$
The product in the rank formula is $1\cdot4=4$, not $1\cdot2$; all four classes on the companion curve are essential.
For the second curve, $a=5,b=-6$, and
$$
E':Y^2=X(X^2-10X+49).
$$
The candidate classes on $E$ are $\{\pm1,\pm2,\pm3,\pm6\}$. The identity, $T$, $(2,4)$ and $(-3,6)$ show that
$$
\{1,-6,2,-3\}\subseteq A.
$$
This is a subgroup of order $4$. Its other coset is $\{3,6,-1,-2\}$, so it suffices to exclude the representative $3$.
The <modulo-eight obstruction to a two-isogeny descent class> uses the corresponding <two-isogeny descent quartic>,
$$
N^2=3U^4+5U^2V^2-2V^4,\qquad\gcd(U,V)=1.
$$
If both $U,V$ are odd, its right side is $6$ modulo $8$. If $U$ is odd and $V$ even, it is $3$ or $7$ modulo $8$. If $U$ is even and $V$ odd, it is $6$ or $2$ modulo $8$. These are all primitive parity cases, and none is a square modulo $8$. Hence the class $3$ is impossible. Since $A$ is a subgroup, every class in its coset is impossible, and therefore
$$
\boxed{A=\{1,2,-3,-6\},\qquad |A|=4.}
$$
On the companion curve, the only candidates are $\pm1,\pm7$. Its quadratic factor is $(X-5)^2+24>0$, so every nonzero real affine $X$ is positive. The exceptional value $[49]$ is $1$, and the point $(7,14)$ supplies the class $7$. Thus
$$
\boxed{A'=\{1,7\},\qquad |A'|=2.}
$$
The rank formula gives $2^r=4\cdot2/4=2$, so
$$
\boxed{\operatorname{rank}\bigl(y^2=x(x^2+5x-6)\bigr)=1.}
$$
Every included class has an explicit rational representative and every excluded coset has a proved real or congruence obstruction, so these are exact ranks rather than bounds obtained from a point search.
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