= Solution
Write $s=\sigma+it$ and let $\rho=\beta+i\gamma$ run over <Nontrivial zeros of the Riemann zeta function>, including <multiplicity>. The bounds involving $\log|t|$ below are for $|t|\ge3$. At bounded heights their uniform versions use $\log(|t|+3)$; $\log|t|$ itself cannot be the right bound near <zero> or $|t|=1$.
Start with the <Hadamard factorization> of the <Riemann xi function>, an <entire function> of order one. Its <logarithmic derivative> has the convergent expansion
$$
\frac{\xi'}{\xi}(s)=B+\sum_\rho\left(\frac1{s-\rho}+\frac1\rho\right).
$$
Differentiating $\xi(s)=\tfrac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)$ gives
$$
\frac{\zeta'}{\zeta}(s)
=B+\sum_\rho\left(\frac1{s-\rho}+\frac1\rho\right)
-\frac1s-\frac1{s-1}+\frac12\log\pi-\frac12\frac{\Gamma'}{\Gamma}(s/2).
$$
The real sum of $1/\rho$ converges absolutely: $0\le\beta\le1$ makes its terms $O(\gamma^{-2})$ at large height, and the order-one <Hadamard factorization> ensures the required summability. Thus these terms and $\Re B$ form a fixed constant. The <Stirling formula> for the <digamma function> gives, uniformly for $1<\sigma\le2$ and $|t|\ge3$,
$$
F(\sigma+it):=-\Re\frac{\zeta'}{\zeta}(\sigma+it)
=\frac12\log|t|-\sum_\rho\frac{\sigma-\beta}{(\sigma-\beta)^2+(t-\gamma)^2}+O(1).
$$
This is the useful real form of the <global partial-fraction expansion of the zeta logarithmic derivative>. Every term subtracted is nonnegative because the <Euler product> excludes <zeros> with $\beta>1$. At the simple <pole> $s=1$, one also has $F(\sigma)=1/(\sigma-1)+O(1)$ as $\sigma\downarrow1$.
The <three-four-one zero-free-region argument> now detects a <zero> too close to one. The <Von Mangoldt function> is nonnegative and the absolutely convergent <Dirichlet series> gives
$$
0\le3F(\sigma)+4F(\sigma+it)+F(\sigma+2it),
$$
because the corresponding coefficient at $n$ is $\Lambda(n)n^{-\sigma}[3+4\cos(t\log n)+\cos(2t\log n)]$. If $\beta+it$ is a <zero>, retain its term in the expansion at height $t$ and discard the other nonpositive terms. The expansion at height $2t$ gives an upper bound even without retaining a particular <zero>. We obtain, for an absolute constant $C_0>0$,
$$
\frac4{\sigma-\beta}\le\frac3{\sigma-1}+C_0\log|t|.
$$
Choose $0<a<1$ so small that $5/(9a)>C_0$, put $b=a/8$, and take $\sigma=1+a/\log|t|$. If $1-\beta\le b/\log|t|$, the last inequality would require
$$
\frac{32}{9a}\le\frac3a+C_0,
\qquad\text{or}\qquad\frac5{9a}\le C_0,
$$
a contradiction. Therefore
$$
\boxed{\zeta(\sigma+it)\ne0\quad\text{for}\quad
\sigma>1-\frac b{\log|t|},\quad |t|\ge3,}
$$
for some fixed $b>0$. This proves the required <zero-free region of the Riemann zeta function>. Nonvanishing on $\Re s=1$ at bounded nonzero heights, proved in Question 2, and the isolation of <zeros> allow a decrease of $b$ giving the uniform formulation with $\log(|t|+3)$ as well. Near $s=1$, apply this observation to the nonzero <holomorphic function> $(s-1)\zeta(s)$.
Next evaluate the real expansion at $\sigma=2$. The absolutely convergent <Dirichlet series> bounds $|\zeta'/\zeta(2+it)|\le\sum_n\Lambda(n)n^{-2}$ uniformly in $t$, so
$$
\sum_\rho\frac{2-\beta}{(2-\beta)^2+(t-\gamma)^2}
=\frac12\log|t|+O(1).
$$
Since $1\le2-\beta\le2$, each term on the left is at least $1/[4+(t-\gamma)^2]$. The <smoothed zeta zero-count bound> follows:
$$
\boxed{\sum_\rho\frac1{4+(t-\gamma)^2}\ll\log|t|.}
$$
Each <zero> satisfying $|t-\gamma|<1$ contributes more than $1/5$, so this also gives the <local zero count for the Riemann zeta function>:
$$
\boxed{\#\{\rho:|t-\Im\rho|<1\}\ll\log|t|.}
$$
The same kernel sum is bounded at compact heights, since its tail converges uniformly there; the bounds with $\log(|t|+3)$ are consequently valid for all real $t$.
For completeness, the contour estimate used in Question 2 follows from these very formulas. Subtract the complex <logarithmic derivative> expansion at $2+it$ from that at $\sigma+it$, with $1-b/(2\log(T+3))\le\sigma\le2$ and $|t|\le T$. Terms with $|t-\gamma|>1$ have differences $O((t-\gamma)^{-2})$, whose sum is $O(\log(|t|+3))$ by the kernel bound. There are $O(\log(|t|+3))$ remaining terms. After reducing $b$, the <zero-free region of the Riemann zeta function> separates every such <zero> horizontally from the half-width boundary by $\gg1/\log(T+3)$: their heights satisfy $|\gamma|\le T+1$. Thus on the left edge, and on the horizontal edges at $t=\pm T$, each reciprocal is $O(\log(T+3))$. The <Gamma function> and <pole> terms satisfy the same required bound; at small heights on the left edge the <pole> distance is $\gg1/\log(T+3)$. This proves $|\zeta'/\zeta|\ll\log^2(T+3)$ on those edges and justifies the contour shift without choosing heights that dodge unknown <zeros>.
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