= Solution
For $E:y^2=x^3-x^2+25x$, the addition formulas give
$$
P+T=(25,-125),
\qquad
Q+T=(5,-15)=-Q,
\qquad
P+Q=(5/4,-45/8).
$$
In particular, $[2]Q=T$ and $Q$ has order four, while the nonintegral coordinate of $[2]P=(144/25,-2172/125)$ and part (a) show that $P$ has infinite order.
For <two-isogeny descent>, write
$$
E:y^2=x^3+ax^2+bx,
\qquad
E':y^2=x^3-2ax^2+(a^2-4b)x.
$$
The rational point $(0,0)$ is the kernel of a degree-two isogeny $E\to E'$. The square-class maps send an affine point with $x\ne0$ to $[x]\in\mathbb Q^*/\mathbb Q^{*2}$, send $O$ to $1$, and send $(0,0)$ to $[b]$ or $[a^2-4b]$ on the two curves. Their images determine the rank through
$$
2^r=\frac{|\alpha(E(\mathbb Q))|\,|\alpha'(E'(\mathbb Q))|}{4}.
$$
Here
$$
E':y^2=x^3+2x^2-99x.
$$
On $E$, the possible square classes divide $25$; real solubility excludes the negative classes, while $P$ and $Q$ exhibit $1$ and $5$. Thus $\alpha(E(\mathbb Q))=\{1,5\}$. On $E'$, the points
$$
(0,0),\quad(-1,10),\quad(11,22)
$$
exhibit the classes $-11,-1,11$, along with $1$. The remaining candidate classes are divisible by $3$. Their homogeneous spaces
$$
N^2=dU^4+2U^2V^2-\frac{99}{d}V^4
$$
have no primitive solution modulo $9$: reduction modulo $3$ first forces $3\mid UV$, and then the equation is congruent to $3$ or $6$ modulo $9$. Hence
$$
\alpha'(E'(\mathbb Q))=\{1,-1,11,-11\}.
$$
The rank formula gives $2^r=2\cdot4/4=2$, so $r=1$.
At the good primes $7$ and $19$, direct counts give
$$
\#E(\mathbb F_7)=12,
\qquad
\#E(\mathbb F_{19})=16.
$$
The <reduction of torsion points on an elliptic curve> injects rational torsion into both groups, so its order divides $\gcd(12,16)=4$. Since $Q$ has order four, the torsion subgroup is $\mathbb Z/4\mathbb Z$. Therefore
$$
E(\mathbb Q)\cong\mathbb Z/4\mathbb Z\times\mathbb Z,
$$
so $t=1$, $d_1=4$, and $r=1$.
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