= Solution
In the position representation the <Time-dependent Schrödinger equation> is
$$
\boxed{i\hbar\partial_t\psi=-\frac{\hbar^2}{2m}\partial_x^2\psi+V(x)\psi.}
$$
For a normalized state define <probability density> and <probability current> by
$$
\rho=|\psi|^2,\qquad
j=\frac{\hbar}{2mi}(\overline\psi\psi_x-\psi\overline{\psi_x})
=\frac\hbar m\operatorname{Im}(\overline\psi\psi_x).
$$
Multiplying the equation by $\overline\psi$ and its <complex conjugate> by $\psi$, then subtracting, cancels the real potential. The resulting identity is
$$
\partial_t|\psi|^2=\frac{i\hbar}{2m}
(\overline\psi\psi_{xx}-\psi\overline{\psi_{xx}})=-\partial_xj.
$$
Thus $\boxed{\partial_t\rho+\partial_xj=0}$. Integrating over an interval gives $d\int_b^c\rho\,dx/dt=j(b)-j(c)$: probability changes only through flux across the endpoints. This is local <conservation of quantum probability>.
Continuity with the zero exterior imposes zero endpoint values on the <infinite square well>. Solving the stationary equation with these <boundary conditions> gives
$$
\boxed{E_n=\frac{n^2\pi^2\hbar^2}{2ma^2},\qquad
\Psi_n(x)=\sqrt{\frac2a}\sin\frac{n\pi x}a,\quad n\geq1,}
$$
with zero extension outside. The <integral> of $\sin^2(n\pi x/a)$ is $a/2$, which fixes the <normalization>. The initial coefficients are $3/5$ and $4/5$, whose squared moduli sum to one. Each energy component acquires its stationary phase:
$$
\boxed{\psi(x,t)=\frac35\Psi_1(x)e^{-iE_1t/\hbar}
+\frac45\Psi_2(x)e^{-iE_2t/\hbar}.}
$$
At the midpoint, $\Psi_1=\sqrt{2/a}$, $\Psi_1'=0$, $\Psi_2=0$, $\Psi_2'=-(2\pi/a)\sqrt{2/a}$. Hence
$$
\overline\psi\psi_x=-\frac{48\pi}{25a^2}e^{-i(E_2-E_1)t/\hbar}.
$$
Taking its imaginary part proves
$$
\boxed{j(a/2,t)=\frac{48\pi\hbar}{25ma^2}
\sin\left(\frac{3\pi^2\hbar}{2ma^2}t\right),\qquad
A=\frac{48\pi\hbar}{25ma^2},\quad\omega=\frac{3\pi^2\hbar}{2ma^2}.}
$$
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