Let $T:V\to V$ be a self-adjoint operator on a Hilbert space $V$. We know that its spectrum $\sigma(T)$ is contained in some finite interval $[a,b]$. We consider the ring homomorphism \begin{align*} \varphi : &\Bbb{R}[X] \to \Bbb{R}[T] \ \text{given by} \ f:=f(X) \mapsto f(T), \end{align*} where $\Bbb R[X]$ denotes the ring of all polynomial functions on $[a,b]$. We also know that $\varphi$ is uniformly continuous with the sup-norm on $\Bbb{R} [X]$ and the operator-norm on $\Bbb {R}[T]$. So we can extend the map $\varphi$ to a continuous map $$\overline{\varphi}:\overline{\Bbb{R}[X]}\to\overline{\Bbb{R}[T]},$$ where $\overline{\Bbb R[X]}$ is the sup-norm closure of $\Bbb{R}[X]$ in $C([a,b])$, and the latter is the operator-norm completion of $\Bbb{R}[T]$. Note that by Weierstrass Approximation Theorem, we have $\overline{\Bbb{R}[X]}=C([a,b])$ where $C([a,b])$ is the ring of real-valued continuous functions on $[a,b]$.
My question 1.: Show that if $f\in C([a,b])$ and $f(x_1)=1$ for some $x_1\in[a,b]$ then $f(T)\neq 0$.
My question 2. Let $\tau(T)\subseteq[a,b]$ be the simultaneous zero-set of $\operatorname{Ker}(\overline{\varphi})$. How can we show that if $f\geq 0$ on $\tau(T)$ then $f(T)\geq 0$.
In fact, from the following notes
http://www-users.math.umn.edu/~garrett/m/fun/notes_2016-17/spectral_theorem_bdd.pdf
we know that if $f\geq 0$ on $[a,b]$ then $f(T)\geq 0$ at the first page.
Its not true as you have written it. Consider $T=\Bbb1$, take $[a,b]=[0,1]$ and look at $f(X)=1-X$. Clearly $f(0)=1$ but $f(T)=0$.
The statement can be modified to be true however:
If you know the Gelfand-Naimark theorem then the map $C([a,b])\to\Bbb C[T]$ factorises over $$C[a,b]\to C(\sigma(T))\overset{\cong}\to \Bbb C[T]$$ and it is clear that $f$ is not sent to zero in the first map. If you don't know this theorem we can also do it on foot:
Let $p_n\to f$, $f(T)$ is defined as $\lim_n p_n(T)$. Since $f(x_1)\neq0$ we can assume |$p_n(x_1)|$ is always larger than some number $c$. Since $$\sigma( p_n(T) )=\{ p_n(x)\mid x\in \sigma(T)\}$$ you have that $$\|p_n(T)\|=\sup_{x\in \sigma(p_n(T))}|x|=\sup_{x\in\sigma(T)}|p_n(x)|≥c$$ it follows that $\lim_n p_n(T)$ cannot be $0$.