= Solution
Inside the <infinite square well>, the time-independent <Schrödinger equation> is $-\hbar^2u''/(2m)=Eu$ with $u(-a)=u(a)=0$. For $E<0$ the exponential solutions cannot satisfy both conditions except trivially, and for $E=0$ neither can a linear function. For $E>0$ put $k=\sqrt{2mE}/\hbar$. The first boundary condition selects $u=C\sin(k(x+a))$, and the second gives $\sin(2ak)=0$. Thus
$$
\boxed{u_n(x)=\frac1{\sqrt a}\sin\!\left(\frac{n\pi(x+a)}{2a}\right),\qquad
E_n=\frac{\hbar^2\pi^2n^2}{8ma^2},\quad n=1,2,\ldots}
$$
on $[-a,a]$, with $u_n=0$ outside. Their squared integrals are one, and the sine <orthogonality> identity gives an <orthonormal set>. The sine <Fourier basis> on the interval is complete: odd extension to an interval of length $4a$ reduces this to completeness of the <Fourier series> basis. Hence these are a complete set of normalized <energy eigenstates>.
Choosing the harmless sign of the second <energy eigenstate> so that $v_1=a^{-1/2}\cos(\pi x/(2a))$ and $v_2=a^{-1/2}\sin(\pi x/a)$, the specified <wave function> is $v_1/\sqrt5+2v_2/\sqrt5$. The <Born rule> therefore gives
$$
\boxed{\Pr(E=E_1)=\frac15,\qquad\Pr(E=E_2)=\frac45},
$$
with no other possible measured energies. The two <energy eigenvalues> are $E_1=\hbar^2\pi^2/(8ma^2)$ and $E_2=4E_1$.
After the first measurement gives the <ground state>, the old state is $v_1$. Doubling the width puts the walls at $\pm2a$, and the new normalized <ground state> is $w_1=(2a)^{-1/2}\cos(\pi x/(4a))$. The stipulated unchanged state is zero outside the original interval, so the <probability amplitude> is
$$
\langle w_1,v_1\rangle
=\frac1{\sqrt2\,a}\int_{-a}^a
\cos\!\left(\frac{\pi x}{4a}\right)\cos\!\left(\frac{\pi x}{2a}\right)\,dx
=\frac8{3\pi}.
$$
Squaring gives
$$
\boxed{\Pr(\text{new ground state}\mid\text{old ground state})=\frac{64}{9\pi^2}}.
$$
The <ground-state overlap after sudden expansion of a square well> stays unchanged during subsequent evolution in the fixed new well, because the energy-basis coefficients acquire only phases. The answer is conditional on the already observed old <ground state>; its earlier probability $1/5$ is not another factor in this conditional question.
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