= Solution
We first determine the torsion. The rational <2-torsion> points are $O,(0,0),(5,0),(-5,0)$. At the good primes three and seven, part (b) gives group orders four and eight. Prime-to-residue-characteristic torsion injects under reduction: an element in the kernel has a formal parameter $t$ of positive valuation, and multiplication by an integer prime to the residue characteristic has series $mt+O(t^2)$ with unit leading coefficient, so cannot kill a nonzero such parameter. At seven this excludes rational 3-primary torsion; at three it excludes 7-primary torsion and every odd-primary component other than the already excluded 3-primary component. Thus all torsion is 2-primary and injects at three, bounding its order by four. Consequently
$$
E_5(\mathbb Q)_{\mathrm{tors}}=\{O,(0,0),(5,0),(-5,0)\}\cong(\mathbb Z/2\mathbb Z)^2.
$$
For the rank, use <two-isogeny descent> with
$$
E:y^2=x^3-25x,\qquad E':Y^2=X^3+100X.
$$
The <two-isogeny formula> gives
$$
\phi(x,y)=\left(x-\frac{25}{x},\ y\left(1+\frac{25}{x^2}\right)\right),\qquad
\psi(X,Y)=\left(\frac{Y^2}{4X^2},\ \frac{Y}{8}\left(1-\frac{100}{X^2}\right)\right).
$$
They extend at the exceptional points, have kernels $\{O,(0,0)\}$ on their respective curves, and satisfy $\psi\phi=[2]$. These assertions can be checked by substitution; for example the composed x-coordinate is $(x^2+25)^2/[4x(x^2-25)]$, the duplication x-coordinate. The maps respect addition because they are morphisms fixing $O$: pushforward of <divisor classes> is additive, takes <principal divisors> to <principal divisors> by the norm of a rational function, and under the identification in part 1(a) sends $[(P)-(O)]$ to $[(\phi P)-(O)]$, and likewise for $\psi$.
Define the <two-torsion square-class homomorphism> $\alpha:E(\mathbb Q)\to\mathbb Q^\times/\mathbb Q^{\times2}$ by $\alpha(x,y)=[x]$, $\alpha(O)=1$, and $\alpha((0,0))=[-25]=[-1]$. Define $\alpha'$ on $E'$ similarly, with $\alpha'((0,0))=[100]=1$. To see the homomorphism property, a nonvertical line $y=mx+c$ cuts either curve in three points whose x-coordinates have product $c^2$. The chord-and-tangent law therefore makes their square classes multiply to one. When the line passes through $(0,0)$, the two remaining x-coordinates have product equal to the coefficient of $x$, exactly explaining the assigned value there. Vertical lines give the inverse-point relation. The same argument includes repeated tangent intersections.
The kernels are $\ker\alpha=\psi E'(\mathbb Q)$ and $\ker\alpha'=\phi E(\mathbb Q)$. Here is an explicit check, not just an invocation of the descent correspondence. The x-coordinate of every nonexceptional $\psi$-image is a square. Conversely, if $(x,y)\in E$ has $x=s^2\ne0$, put $X=2x+2y/s$ and $Y=2sX$. Then $X^2-4xX+100=0$, so $(X,Y)\in E'$ and the displayed $\psi$ maps it to $(x,y)$. For $E'$, every nonexceptional $\phi$-image has $X=(y/x)^2$; if $X=s^2\ne0$, put $x=(X+Y/s)/2$, $y=sx$. Substitution gives a point of $E$ mapping to $(X,Y)$. The point $(0,0)\in E'$ has square class one and equals $\phi(5,0)$; the point $(0,0)\in E$ has nonsquare class $-1$. Together with the origins these checks handle every exception.
The <prime-support bound in two-isogeny descent> can be read directly from $y^2=x(x^2+B)$. At a prime not dividing the integral coefficient $B$, positive valuation of $x$ gives $2v(y)=v(x)$, while negative valuation gives $2v(y)=3v(x)$; either way the valuation of $x$ is even. Thus the first image has only the candidate classes $1,-1,5,-5$. All four occur, already at the four torsion points. Hence $\#\alpha(E)=4$.
On $E'$, nonzero real x-coordinates are positive since $X(X^2+100)$ has the sign of $X$. Prime support leaves only $1,2,5,10$. The point $(5,25)$ supplies class five. Class two would give the <quartic covering in a two-isogeny descent>
$$
W^2=2U^4+50V^4,\qquad\gcd(U,V)=1.
$$
Indeed write $X=2(U/V)^2$ and clear the square factor in the curve equation. If $5\nmid U$, the right side is two modulo five, impossible for a square. If $5\mid U$, then $5\nmid V$ and $5\mid W$; division by twenty-five gives $(W/5)^2\equiv2V^4\equiv2\pmod5$, again impossible. This is the <five-adic obstruction to square class two on the curve with coefficient one hundred>. Because the image is a subgroup and contains five, class ten is excluded as well. Therefore $\alpha'(E')=\{1,5\}$ has size two.
Now $[E:\psi E']=4$. The quotient $\psi E'/2E$ is isomorphic to $E'/\phi E$: the kernel of the induced surjection is $\ker\psi/(\ker\psi\cap\phi E)$, which is trivial because $(0,0)\in E'$ equals $\phi(5,0)$. Thus $[E:2E]=4\cdot2=8$. Finite generation follows from the <Mordell-Weil theorem>, whose height-descent proof is supplied in the essay below. A finitely generated group with the torsion already found has $[E:2E]=2^{r+2}$, so $r=1$. Hence
$$
\boxed{E_5(\mathbb Q)\cong(\mathbb Z/2\mathbb Z)^2\times\mathbb Z.}
$$
The point $(-4,6)$ has infinite order, since it is not one of the torsion points. No assertion that this particular point generates the free factor is needed.
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