I am following the proof of the following theorem in Reed & Simon's book on functional analysis:
Let $\{A_n\}_{n=1}^\infty$ and $A$ be self-adjoint operators and suppose that $A_n \rightarrow A$ in the norm-resolvent sense. Then if $\mu \not\in \sigma(A)$ then $\mu \not\in\sigma(A_n)$ for $n$ sufficiently large and $$\|R_\mu(A_n) - R_\mu(A) \| \rightarrow 0.$$
Their proof begins as
We need only consider the case where $\mu$ is real. Since $\mu \in \rho(A)$, there is a $\delta > 0$ so that $(\mu - \delta, \mu + \delta) \cap \sigma(A) = \emptyset$. Thus, by the functional calculus, $$\|R_{\mu + i\delta/3}(A)\| < \frac1\delta.$$
I am confused on this step, how are they using the functional calculus to obtain this bound?
Similarly in another proof they state that
...the statement that $(a,b) \cap \sigma(A) = \emptyset$ is equivalent to the statement that $$\|(A - \lambda_0)^{-1}\| \leq \frac{\sqrt{2}}{b-a}$$ where $$\lambda_0 = \frac{a+b}{2} + i\big(\frac{b-a}{2}\big).$$
How are they using the functional calculus to obtain these bounds?
A tad late but I just worked these out myself last week so I might as well write up an answer.
For the first one:
Let's denote by $r_z$ for $z \in \mathbb{C} \setminus \mathbb{R}$ the function $x \mapsto \frac{1}{z - x}$, so that $R_z(A) = r_z(A)$, resolvent on the left, functional calculus on the right, and by $m_z$ the function $x \mapsto |r_z(x)|^2 = \frac{1}{|z - x|^2}$.
We know that $\|R_{\mu + i\delta/3}(A)\| = \|r_{\mu + i\delta/3}(A)\| = \|r_{\mu + i\delta/3}\|_{\infty, \sigma(A)}$. The next part is only there for rigor, but it's intuitive that $r_z$ attains its maximum modulus at $\operatorname{Re}(z)$ and is strictly monotonic left and right of said maximum, hence in this case, since we know that $(\mu - \delta, \mu + \delta) \cap \sigma(A) = \emptyset$, we'll only be concerned with the (absolute) values of $r_{\mu + i\delta/3}$ at $\mu - \delta$ and $\mu + \delta$.
We have: $$m_z(x) = \frac{1}{\left(x - \operatorname{Re}(z)\right)^2 + \operatorname{Im}(z)^2}$$ This is differentiable on $\mathbb{R}$, with: $$\forall x \in \mathbb{R},\quad m_z'(x) = \frac{2\left(\operatorname{Re}(z) - x\right)}{\left(\left(x - \operatorname{Re}(z)\right)^2 + \operatorname{Im}(z)^2\right)^2}$$ Therefore $m_z$ is increasing until $\operatorname{Re}(z)$ and decreasing after that point, hence implying what was mentioned above.
Moreover: $$\frac{1}{\left|\mu + i\delta/3 - (\mu \pm \delta)\right|^2} = \frac{1}{\left|\mp \delta + i\delta/3 \right|^2} = \frac{1}{\frac{10}{9} \delta^2} < \frac{1}{\delta^2}$$ which nets us the desired estimate: $$\|R_{\mu + i\delta/3}(A)\| < \frac{1}{\delta}$$
The same can then be achieved in the second part of your question, replacing $\mu + i\frac{\delta}{3}$, $\mu - \delta$ and $\mu + \delta$ with $\lambda_0$, $a$ and $b$ respectively.
The key factor is that $m_{\lambda_0}$ is strictly monotonic left and right of $\operatorname{Re}(\lambda_0) = \frac{a + b}{2}$, hence only (up to) two points $c_+(\alpha)$ and $c_-(\alpha)$ are the preimages to any given value $\alpha$ of $m_{\lambda_0}$. If you know that $\|R(\lambda_0, A)\| = \|r_{\lambda_0}\|_{\infty, \sigma(A)} = \alpha$ then you now know that $c_-(\alpha)$ or $c_+(\alpha)$ belongs to $\sigma(A)$ and also that $(c_-(\alpha), c_+(\alpha))$ lies in $\rho(A)$.
It just so happens that we have the following: $$m_{\lambda_0}(a) = m_{\lambda_0}(b) = \frac{1}{\left|\pm \frac{b - a}{2} + i\frac{b-a}{2}\right|^2} = \frac{2}{(b - a)^2}$$ hence the result.
I was stuck just like you when I first saw the proofs mentioning only "by the functional calculus", but I had another proof method for the latter point using knowledge about power series and some amount of geometry which make this have more intuitive sense, at least for me. Showcasing this proof will be the objective of the rest of this here answer. Consider this a bonus.
Before moving on to the second part of the question but done with geometry, I'll address something that'll be useful in regard to the norm of the resolvent, though moreso for the second aspect. You can go about what I'll say differently (for example it's probably done in another way somewhere in or throughout the same volume of Reed and Simon the question is based on), this is just my preferred way to see things.
As you should know, at least for a self-adjoint operator, $R(z,A)^* = R(\bar{z}, A)$, and $R(z,A)$ and $R(\bar{z}, A)$ commute, thus $R(z,A)$ is a normal operator for all $z \in \rho(A)$. This provides the equality: $$\left\|R(z,A)^n\right\| = {\|R(z,A)\|}^n$$ for all $n \geq 1$ (see for example Toshio Kato's Perturbation Theory for Linear Operators Chapter 1, that chapter deals with finite-dimensional spaces but the proof of that statement works for all normal operators from all Hilbert spaces. Alternatively, there's also the answer to the post Is $\Vert A^n \Vert = \Vert A \Vert^n$ for normal operator $A$ on inner product space?, and probably a lot of other posts on MSE I haven't touched upon).
Now, the function $z \mapsto R(z,A)$ is one of these functions $f$ for which, if their Taylor series at some $z_0$ converges at some point $z_1$, it has to converge to (an analytic extension of) $f$ at said point $z_1$ (this might seem like a trivial thing to have, but consider for example the function $z \in \mathbb{C} \setminus \mathbb{R} \mapsto \begin{cases} 0 \text{ if } \operatorname{Im}(z) > 0 \\ 1 \text{ if }\operatorname{Im}(z) < 0\end{cases}$.)
This is because, if the series $\sum_{n=0}^\infty (-u)^n R(z,A)^{n+1}$ converges, then it is just a matter of composing by $(z+u)\operatorname{Id} - A$ (with a few justifications for commuting sum and composition, etc...) to establish that $z+u \in \rho(A)$ with $R(z+u, A) = \sum_{n=0}^\infty (-u)^n R(z,A)^{n+1}$.
This implies that, combined with the fact that the radius of the Taylor series of a holomorphic function $f : \Omega \to \mathbb{C}$ at $z_0 \in \Omega$ is at least $\operatorname{dist}(z_0, \partial \Omega)$ thanks to how you prove that a holomorphic function is representable by power series (for the emphasis on "at least", remember the example from earlier, though even for connected domains think of an $f$ with a removable singularity), we get that: $$\mathrm{radius}\left(\sum_{n=0}^\infty (-u)^n R(z,A)^{n+1}\right) = \operatorname{dist}\left(z, \partial \rho(A)\right) = \operatorname{dist}\left(z, \sigma(A)\right)$$
However, the radius of convergence of the power series $\sum_{n=0}^\infty (-u)^n R(z, A)^{n+1}$ is, as per Hadamard's formula, the inverse of $\limsup_{n \to \infty} \|R(z,A)^{n+1}\|^{\frac{1}{n}}$. We obtain from the prior observation that: $$\left\|R(z,A)^{n+1}\right\|^{\frac{1}{n}} = \|R(z,A)\|^{\frac{n+1}{n}} \xrightarrow[n \to \infty]{} \|R(z,A)\|$$ hence the radius of the Taylor series of $R(\cdot, A)$ at $z$ is precisely $\|R(z,A)\|^{-1}$.
This can help clarify what happens after the estimate with the $\frac{1}{\delta}$ in the proof the first part of your post is extracted from, but it's not like you need any of what I've just explained to understand that passage. Note that the convergence to $0$ of $\|R(\mu, A_n) - R_\mu(A)\|$ is not really automatic as norm resolvent convergence only uses resolvents $R(z,\cdot)$ with non-real $z$, but with Theorem VIII.20(a) and an adequate function $r_{\mu,\delta}$ taking the role of $r_\mu$ you can use the functional calculus to obtain that convergence.
Onto the actual geometric proof: the image at the bottom shows how you get the $\frac{b-a}{\sqrt{2}}$.
The idea is that $(a,b)$ lies in $\rho(A)$ if and only if the open disk of center $\lambda_0$ whose circle passes by $a$ and $b$ lies in $\rho(A)$. The radius of that disk is then obtained by the usual hypotenuse formula for a right triangle, using the triangle passing by $\lambda_0$, $a$ and $\frac{a+b}{2}$ whose two other sides' lengths are both $\frac{b-a}{2}$, hence the $\frac{b - a}{\sqrt{2}}$ radius.
But having a disc centered at $\lambda_0$ lying in $\rho(A)$ is equivalent to having it be contained in the disc centered at $\lambda_0$ of radius the radius of the Taylor series of $R(\cdot, A)$ at $\lambda_0$, aka $\|R(z,A)\|^{-1}$, since, as was explained previously, said radius is exactly the distance of $\lambda_0$ to $\sigma(A)$ (we wouldn't have been able to conclude that with only an "at least").
Therefore, since inclusion of discs of same center is just an inequality between the radii: $$(a,b) \cap \sigma(A) = \emptyset \Leftrightarrow D\left(\lambda_0, \frac{b-a}{\sqrt{2}}\right) \subseteq D\left(\lambda_0, \|R(z,A)\|^{-1}\right) \Leftrightarrow \|R(z,A)\| \leq \frac{\sqrt{2}}{b-a}$$