= Solution
Assume first the <Riemann hypothesis>. Replace any $x\geq2$ by $y=\lfloor x\rfloor+1/2$; then $\psi(x)=\psi(y)$ and $\langle y\rangle\geq1/2$. Take $T=y$ in part a. Every nontrivial zero has real part $1/2$, and the <Riemann–von Mangoldt formula> implies
$$
\sum_{|\gamma|\leq y}\frac1{|\rho|}\ll(\log y)^2.
$$
Consequently
$$
\sum_{|\gamma|\leq y}\left|\frac{y^\rho}{\rho}\right|
\ll y^{1/2}(\log y)^2,
$$
while both truncation errors in part a are $O((\log y)^2)$. Thus
$$
\psi(x)=x+O\left(x^{1/2}(\log x)^2\right),
$$
which implies the stated $O_\epsilon(x^{1/2+\epsilon})$ estimate.
Conversely, suppose that estimate holds for every $\epsilon>0$. For $\Re s>1$, <partial summation> gives
$$
-\frac{\zeta'(s)}{\zeta(s)}
=s\int_1^\infty\psi(x)x^{-s-1}\,dx
=\frac{s}{s-1}
+s\int_1^\infty(\psi(x)-x)x^{-s-1}\,dx.
$$
Given any $s$ with $\Re s>1/2$, choose $\epsilon<\Re s-1/2$. The error hypothesis makes the last integral locally uniformly convergent there, so it supplies a holomorphic continuation of
$$
-\frac{\zeta'(s)}{\zeta(s)}-\frac{s}{s-1}
$$
to the half-plane $\Re s>1/2$. A zero of $\zeta$ in that half-plane would create a pole of its <logarithmic derivative>, so none exists. The <Functional equation of the Riemann zeta function> reflects every nontrivial zero with real part below $1/2$ to one above $1/2$. All nontrivial zeros must therefore lie on the <critical line>, proving the <Riemann hypothesis equivalence for the second Chebyshev function>.
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