Let $\mathscr{C}$ denote the collection of all finite intersections $\bigcap_{i=1}^{n}\{X_{i}\leq x_{i}\}$, where $x_{1},...,x_{n}\in\mathbb{R}$ and $n\geq 1$, and let $\mathscr{B}(X)$ denote the Borel $\sigma$-algebra of a topological pair $(X,\tau)$.
I will point the main ideas but some details will be left for you to verify.
(1.) The first question asks if two $\sigma$-algebras are equal, namely $\sigma(\tilde{X})$ and $\sigma(\mathscr{C})$. For this it suffices to show that the generator sets of $\sigma(\tilde{X})$ are in $\sigma(\mathscr{C})$, whence $\sigma(\tilde{X})\subset \sigma(\mathscr{C})$, and on the other hand that $\mathscr{C}\subset \sigma(\tilde{X})$, whence $\sigma(\mathscr{C})\subset \sigma(\tilde{X})$.
To start with, note that $\mathscr{B}(\mathbb{R})$ is generated by sets of the form $(-\infty,a]$, where $a\in\mathbb{R}$. So an arbitrary generator of $\sigma(\tilde{X})$ is $\{X_{n}\leq a\}$ for some $n\in\mathbb{N}$ and $a\in\mathbb{R}$. We start by showing that such set lies in $\sigma(\mathscr{C})$. Observe that \begin{equation*} \{X_{n}\leq a\}=\bigcup_{k=1}^{\infty}\left((\{X_{n}\leq a\})\cap(\bigcap_{i=1}^{n-1}\{X_{i}\leq k\})\right). \end{equation*} Since if $\omega\in\{X_{n}\leq a\}$, then choose $k\in\mathbb{N}$ so that $k>\max_{1\leq i \leq n-1}X_{i}(\omega)$, whence $\omega\in\{X_{i}\leq k\}$ for all $i\in\{1,...,n-1\}$. Using this $k$ we obtain the inclusion $\subset$ in the above equation. The inclusion $\supset$ is trivial. Hence $\{X_{n}\leq a\}$ is a countable union of members from $\sigma(\mathscr{C})$, which shows that $\{X_{n}\leq a\}\in\sigma(\mathscr{C})$. Moreover, $\sigma(\tilde{X})\subset \sigma(\mathscr{C})$.
For the converse inclusion, since sets of type $\{X_{n}\leq a\}$ for $n\in\mathbb{N}$ and $a\in\mathbb{R}$ generate $\sigma(\tilde{X})$, then every member of $\mathscr{C}$ is a finite intersection of members from $\sigma(\tilde{X})$, and thus $\sigma(\mathscr{C})\subset \sigma(\tilde{X})$. This concludes that $\sigma(\mathscr{C})= \sigma(\tilde{X})$
(2.) The measurability of $\inf X_{n}$ and $\sup X_{n}$ follows by observing that for all $a\in\mathbb{R}$: \begin{align*}\{\inf X_{n}\leq a\}=\bigcup_{n=1}^{\infty}\{X_{n}\leq a\}\,\,\,\mathrm{and}\,\,\,\{\sup X_{n}\leq a\}=\bigcap_{n=1}^{\infty}\{X_{n}\leq a\}. \end{align*} Hence $\{\inf X_{n}\leq a\},\{\sup X_{n}\leq a\}\in\sigma(\tilde{X})$ for all $a\in\mathbb{R}$.
(3.) From $(2.)$ it follows that $\limsup X_{n}$ and $\liminf X_{n}$ are also measurable, since $\limsup X_{n}=\inf_{n\geq 1}(\sup_{k\geq n} X_{n})$ and $\liminf X_{n}=\sup_{n\geq 1}(\inf_{k\geq n} X_{n})$. Hence if $\lim X_{n}$ exists, then in particular $\lim X_{n}=\liminf X_{n}=\limsup X_{n}$ and it is thus measurable.
For the first part of $(3.)$, let $X(\omega):=\lim X_{n}(\omega)$ (which is either $\limsup X_{n}(\omega)$ or $\liminf X_{n}(\omega)$, as you prefer, and thus measurable) for those $\omega$ for which the limit exists. Since the difference of two measurable functions is measurable, and taking absolute values remains measurable, then the following set \begin{align*} \{\omega\in\Omega:\lim X_{n}(\omega) &=X(\omega)\}=\{\omega\in\Omega:\forall k\in\mathbb{N}\exists m\in\mathbb{N}\,\,\mathrm{s.t.}\,\,\forall n\geq m\,\,|X_{n}(\omega)-X(\omega)|\leq\frac{1}{k}\} \\ &=\bigcap_{k\in\mathbb{N}}\bigcup_{m\in\mathbb{N}}\bigcap_{n\geq m}\{\omega\in\Omega:|X_{n}(\omega)-X(\omega)|\leq\frac{1}{k}\} \end{align*} is as well measurable. I.e. $\{ \lim X_{n}\,\,\mathrm{exists}\}$ is measurable.
I will try to answer all your questions below in the comments section. If there was something that seemed to be false, please make a remark below so I may correct it.