= Solution
\b[Frame definitions and relativistic effects.] Let $S$ be the <inertial frame> of Earth and the stationary asteroid, with $+x$ pointing toward the asteroid. Bob's <inertial frame> $S'$ moves at $+v$ relative to $S$. For two events at Bob's fixed position, $\Delta x'=0$, the inverse <Lorentz transformation> gives
$$
\Delta t=\gamma\Delta t',\qquad\gamma=(1-v^2/c^2)^{-1/2}.
$$
The elapsed <proper time> on Bob's clock is $\Delta\tau=\Delta t'$, so <time dilation> means $\Delta t=\gamma\Delta\tau$: the separation of his clock ticks is larger in Earth coordinates. For a rod at rest in $S'$ with <proper length> $L_0$, a length measurement in $S$ uses simultaneous endpoints, $\Delta t=0$. The direct spatial <Lorentz transformation> gives $L_0=\Delta x'=\gamma\Delta x$, hence <length contraction>:
$$
\boxed{L=L_0/\gamma.}
$$
The simultaneity condition is essential: the endpoint events used to measure length are different from the two clock-tick events used to demonstrate <time dilation>.
\b[Bob's speed.] Food consumption measures Bob's own <proper time>, $\tau_B=4\text{ months}=1/3\,\mathrm{yr}$. The Earth-frame distance is $D=1/4\,\mathrm{ly}$ and the Earth travel <time> is $D/v$. Consequently
$$
\tau_B=\frac{D}{\gamma v},\qquad\gamma\frac vc=\frac{D}{c\tau_B}=\frac34.
$$
Putting $\beta=v/c$, $\beta^2/(1-\beta^2)=9/16$, so
$$
\boxed{v=\frac35c=0.600c,\qquad\gamma=\frac54.}
$$
The Earth travel <time> is $5/12\,\mathrm{yr}=5$ months. Equivalently Bob measures the contracted Earth-asteroid separation $D/\gamma=0.20\,\mathrm{ly}$ and sees the asteroid approach at $0.60c$, taking $1/3\,\mathrm{yr}$ to reach him.
\b[Events A and B.] Set both coordinate origins at Event A. Alice reaches the asteroid at Event B, travelling at $u=2c/3$ in $S$, so
$$
\boxed{\Delta x=\frac14\,\mathrm{ly},\qquad\Delta t=\frac{D}{u}=\frac38\,\mathrm{yr}=4.5\text{ months}.}
$$
The <Lorentz transformations> into Bob's frame give
$$
\begin{aligned}
\Delta t'&=\frac54\left(\frac38-\frac35\frac14\right)\mathrm{yr}=\boxed{\frac9{32}\,\mathrm{yr}=3.375\text{ months}},\\
\Delta x'&=\frac54\left(\frac14-\frac35\frac38\right)\mathrm{ly}=\boxed{\frac1{32}\,\mathrm{ly}=0.03125\,\mathrm{ly}}.
\end{aligned}
$$
Here $c(1\,\mathrm{yr})=1\,\mathrm{ly}$ has been used consistently. With the time-positive sign convention, the <Minkowski interval> is
$$
\begin{aligned}
\Delta s^2&=c^2\Delta t^2-\Delta x^2=\left(\frac9{64}-\frac4{64}\right)\mathrm{ly^2}=\boxed{\frac5{64}\,\mathrm{ly^2}},\\
\Delta s'^2&=c^2\Delta t'^2-\Delta x'^2=\left(\frac{81}{1024}-\frac1{1024}\right)\mathrm{ly^2}=\boxed{\frac5{64}\,\mathrm{ly^2}}.
\end{aligned}
$$
Thus the <Minkowski interval> is invariant. It is timelike, as expected for two events on Alice's subluminal <worldline>. Using the opposite metric sign convention would change both signs together, without changing invariance.
\b[Relative velocity and spacetime diagram.] Divide the differential spatial and temporal <Lorentz transformations> for the same trajectory:
$$
u_x'=\frac{dx'}{dt'}=\frac{\gamma(dx-v\,dt)}{\gamma(dt-v\,dx/c^2)}=\boxed{\frac{u_x-v}{1-vu_x/c^2}}.
$$
For Alice,
$$
\boxed{u'_{\rm Alice}=\frac{(2/3-3/5)c}{1-(2/3)(3/5)}=\frac19c.}
$$
The <spacetime diagram> uses horizontal $x'$ and vertical $ct'$. In these coordinates Bob's <worldline> is $x'=0$, Alice's is $x'=ct'/9$, and the asteroid's is
$$
x'_{\rm asteroid}=\frac D\gamma-vt'=\frac15\,\mathrm{ly}-\frac35ct'.
$$
Thus Alice has slope $d(ct')/dx'=9$, while the asteroid has slope $-5/3$. Their intersection is exactly B, $(x',ct')=(1/32,9/32)\,\mathrm{ly}$. The asteroid later reaches Bob at $(0,1/3)\,\mathrm{ly}$. These intercepts and gradients show the <relativistic velocity addition> geometrically.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021-q7-bob-frame.png]
{title=<Spacetime diagram> in Bob's <inertial frame>, with Alice's <velocity>, asteroid motion and the A-to-B coordinate differences}
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