= Solution
From part (a), $\Theta F=E_2F$. Its product has integer coefficients and begins with $q$, so write $F=\sum_{n\geq1}\tau(n)q^n$, with $\tau(1)=1$. Comparing the coefficient of $q^n$ gives
$$
n\tau(n)=\tau(n)-24\sum_{\ell=1}^{n-1}\sigma_1(\ell)\tau(n-\ell),
$$
which rearranges to
$$
\boxed{(1-n)\tau(n)=24\sum_{\ell=1}^{n-1}\sigma_1(\ell)\tau(n-\ell).}
$$
This is a recurrence for the <Ramanujan tau function>; for example it gives $\tau(2)=-24$ and $\tau(3)=252$.
For the congruence, use the normalized Eisenstein expansions
$$
E_4=1+240\sum_{n\geq1}\sigma_3(n)q^n,\qquad
E_6=1-504\sum_{n\geq1}\sigma_5(n)q^n.
$$
The <modular form> $E_4^3-E_6^2$ is a <cusp form> of weight twelve, and its leading coefficient is $720-(-1008)=1728$. Since $S_{12}$ is one-dimensional and $F$ starts with $q$,
$$
E_4^3-E_6^2=1728F.
$$
Also $D_6E_6$ is a weight-eight form, with constant coefficient $-1/2$. Since $M_8$ is one-dimensional and $E_4^2$ has constant coefficient one, $D_6E_6=-E_4^2/2$, or
$$
2\Theta E_6=E_2E_6-E_4^2.
$$
These identities have rational coefficients whose only displayed denominators are prime to five, so coefficientwise reduction modulo five is legitimate. We have $E_4\equiv1$. Furthermore $d^5\equiv d\pmod5$ gives $\sigma_5(n)\equiv\sigma_1(n)$, while $-504\equiv-24\equiv1$, so $E_6\equiv E_2$. Therefore
$$
F\equiv\frac{1-E_6^2}{3}=\frac{E_6^2-1}{2}\equiv\Theta E_6\pmod5.
$$
The equality in the middle is in $\mathbb F_5$, where $-1/3=1/2$. Comparing coefficients in $\Theta E_6=-504\sum_{n\geq1}n\sigma_5(n)q^n$ gives
$$
\boxed{\tau(n)\equiv n\sigma_5(n)\pmod5.}
$$
This proves the <Ramanujan tau congruence modulo five> from the given small-dimensional modular-form spaces, without assuming that congruence as a theorem.
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