= Solution
Memorylessness means that the ensemble's average <density operator> is $\rho^{(n)}=\pi^{\otimes n}$. Its <spectral decomposition> is
$$
\rho^{(n)}=\sum_{i^n}q_{i_1}\cdots q_{i_n}|\phi_{i^n}\rangle\langle\phi_{i^n}|,\qquad
|\phi_{i^n}\rangle=|\phi_{i_1}\rangle\otimes\cdots\otimes|\phi_{i_n}\rangle.
$$
The product eigenvalues give
$$
S(\rho^{(n)})
=-\sum_{i^n}\left(\prod_jq_{i_j}\right)\log_2\left(\prod_jq_{i_j}\right)
=nH(q)=nS(\pi).
$$
Zero eigenvalues contribute zero to the <Von Neumann entropy> and are excluded from logarithmic typicality tests.
The <quantum typical subspace> is the span of eigenvectors whose eigenvalues satisfy
$$
\left|-\frac1n\log_2(q_{i_1}\cdots q_{i_n})-S(\pi)\right|\leq\varepsilon.
$$
Its orthogonal projection $P_\varepsilon^{(n)}$ selects precisely the classical <typical set> for the eigenvalue distribution $q$. The <typical-set cardinality bounds> and the <weak law of large numbers> therefore give
$$
\boxed{\dim\mathcal T_\varepsilon^{(n)}\leq2^{n(S(\pi)+\varepsilon)}},\qquad
\boxed{\operatorname{Tr}(\rho^{(n)}P_\varepsilon^{(n)})\geq1-\delta}
$$
for any fixed $\delta>0$ and all sufficiently large $n$.
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