= Solution
By part (a), every nontrivial zero $\rho=\beta+i\gamma$ with $|\gamma|\le T$ satisfies $\beta\le1-c_1/\log^9T$ for large $T$, after reducing the positive constant. The <local zero count for the Riemann zeta function> gives
$$
\sum_{|\gamma|\le T}\frac1{|\rho|}\ll1+\sum_{1\le j\le T}\frac{\log(j+3)}j\ll\log^2T.
$$
There are finitely many zeros at bounded height, none at zero or at one, so that part of the sum is bounded. The <truncated explicit formula for the second Chebyshev function> now yields
$$
\boxed{\psi(x)=x+O\bigl(x^{1-c_1/\log^9T}\log^2x\bigr)+O\left(\frac{x\log^2x}T\right),\qquad2\le T\le x.}
$$
Balance the exponent losses $c_1\log x/(\log T)^9$ and $\log T$ by choosing $\log T=(c_1\log x)^{1/10}$. This is the optimal order obtainable from these two errors: making either exponent larger forces the other smaller. Thus, for a positive constant $c_2$,
$$
\boxed{\psi(x)-x\ll x\log^2x\exp\bigl(-c_2(\log x)^{1/10}\bigr).}
$$
The logarithmic prefactor can be absorbed by reducing $c_2$. This is the <prime number theorem error from a logarithmic zero-free region> with ninth-power width.
Under the <Riemann hypothesis>, $|x^\rho|=x^{1/2}$. The same reciprocal-zero sum bounds the zero contribution by $O(x^{1/2}\log^2T)$. Taking $T=x$ makes the truncation error $O(\log^2x)$, so
$$
\boxed{\psi(x)=x+O(x^{1/2}\log^2x).}
$$
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