How to show that $X_n + Y_n \to X + Y$ holds in the $L^1$ norm? Let $X_n$ and $Y_n$ be sequences of random variables. Show that
$X_n + Y_n \to X + Y$ (1)
$X_nY_n \to XY$ (2)
If $\mathbb{P}(X=0) = 0, \;  \frac{Y_n}{X_n} \to \frac{Y}{X}$ (3)
are true for convergence in probability and and convergence almost everywhere, and (1) hold for convergence in $L^1$ aswell.
I have tried the approach with Boole's inequality, interchanging intersections and limits by the monotone convergence theorem, some other stuff, but can't quite seem to get there completely.
As for the $L^1$ convergence, it's my understanding that $X_n \in L^1$ if the sequence is absolutely convergent, and $\int |X_n-X| = 0$.
I remember $L^1$ convergence not being preserved by continuous transformations, and so are there examples where convergence in $L^1$ does not hold for (2) and (3)?

These are standard exercises. So, let me leave behind the ingredients that OP can utilise to formally construct the proofs.
Let $(\Omega, \mathfrak A, \mathbf P) $ be the probability measure space. Let $\langle X_n\rangle_{n\in N},~\langle Y_n\rangle_{n\in \mathbb N}$ be two sequences of extended real valued random variables on a set $D\in\mathfrak A$ converging almost surely on $D$ to the real-valued $X, ~Y$ respectively. That means there exists a null set $N_X: ~X:= \lim X_n(\omega) \in \mathbb R$ for every $\omega\in D\setminus N_X. $ Let $N_Y$ be defined in the same way.
The first two assertions (viz. convergence of sum, product) can be shown readily for all $\omega \in (N_X\cup N_Y) ^\complement.$
Now, let the sequence of extended real-valued random variables $\langle X_n\rangle_{n\in\mathbb N},~\langle Y_n\rangle_{n\in \mathbb N}$ converge in probability measure on $D$ to real-valued random variables $X, ~Y$ i.e. $X_n\overset{\mathbb P}{\rightarrow} X,~  Y_n\overset{\mathbb P}{\rightarrow} Y .$ Then by definition, $\forall \varepsilon >0,$ $$\lim_{n\to\infty}\mathbb P\{D:| (X_n + Y_n) - (X+Y)| \geq \varepsilon\}=0.\tag 1$$
Now, \begin{align}|(X_n + Y_n) -(X+Y) |&\leq |(X_n-X) |+|(Y_n-Y) |\\\implies\{D: |(X_n + Y_n) -(X+Y) |\geq \varepsilon\}&\subset \{D: |(X_n-X) |+|(Y_n-Y) |\geq \varepsilon\}\tag{2.i}\\\wedge~~ \{D: |(X_n-X) |+|(Y_n-Y) |\geq \varepsilon\}& \subset \{D:|(X_n-X) |\geq \varepsilon/2\}\cup \{D:|(Y_n-Y) |\geq \varepsilon/2\}.\tag{2.ii}\end{align}
From $(2) ,$
\begin{align}\mathbf P \{D: |(X_n-X) |+|(Y_n-Y) |\geq \varepsilon\}\leq \mathbf P\{D:|(X_n-X) |\geq \varepsilon/2\}+\mathbf P\{D:|(Y_n-Y) |\geq \varepsilon/2\};\tag 3\end{align}
Apply limits on both sides.
For the product, few results have to be consolidated.
Proposition $1.$ (Riesz) If a sequence of extended real-valued random variables $\langle X_n\rangle_{n\in \mathbb N}$ on $D\in\mathcal A$ converge in probability measure on $D$ to the real-valued $X,$ then there exists a subsequence $\langle X_{n_k}\rangle_{k\in \mathbb N}$ which converges almost surely to $X$ on $D.$
By assumption, for every $\varepsilon> 0, $ there exists $N_\varepsilon\in \mathbb N$ such that $\mathbf P\{D:|X_n -X|\geq \varepsilon\}<\varepsilon$ for $n>N_\varepsilon.$ Take $\varepsilon_k=1/2^k.$ Inductively it can be inferred $\mathbf P\left\{D:|X_{n_k} -X|\geq 1/2^k\right\}<1/2^k$ for $n_k>n_{k-1}.$ Whence $\sum_k \mathbf P\left\{D:|X_{n_k} -X|\geq 1/2^k\right\} <\infty.$ That completes the proof by application of Borel-Cantelli Lemma. $\square$
Observation $1.$ If every subsequence $\langle X_{n_k}\rangle_{k\in\mathbb N}$ of the sequence $\langle X_n\rangle_{n\in\mathbb N}$ has a subsequence $\langle X_{{n_k}_\ell}\rangle_{\ell \in \mathbb N}$ converging almost surely to $X$ on $D,$ then $X_n \overset{\text{a.s.}}{\rightarrow}X.$
Since almost sure convergence implies convergence in probability measure, then every $\langle X_{n_k}\rangle_{k\in\mathbb N}$ of $\langle X_n\rangle_{n\in\mathbb N}$ has a subsequence $\langle X_{{n_k}_\ell}\rangle_{\ell \in \mathbb N}$ which converges in probability measure to $X$ on $D.$ The implication follows. $\square$
To show $X_nY_n \overset{\mathbb P}{\to}XY, $ by Observation $1, $ it suffices to show every subsequence $\langle X_{n_k}Y_{n_k}\rangle_{k\in\mathbb N}$ of thr sequence $\langle X_nY_n\rangle_{n\in \mathbb N}$ has a subsequence $\langle X_{{n_k}_\ell} Y_{{n_k}_\ell}\rangle_{\ell \in \mathbb N}.$
By assumption, the subsequence $\langle X_{n_k}\rangle_{k\in \mathbb N}$ converges in probability measure on $D$ to $X.$ By Theorem $1, $ it is possible to find a subsequence $\langle X_{{n_k}_\ell}\rangle_{\ell \in \mathbb N}$ the converges almost surely on $D$ to $X.$ Similar treatment can be imposed on $\langle Y_n\rangle_{n\in\mathbb N}.$ From the property of almost sure convergence as the product $X_{{n_k}_\ell} Y_{{n_k}_\ell}\overset{\text{a.s.}}{\to}XY;$ since it is a subsequence from an arbitrary subsequence, the rest follows.
Re the ratio, let me rephrase it as $\frac{Y_n}{X_n}\mathbb I(X_n \not= 0).$ Then it is nothing but the application of Continuous Mapping Theorem (cf. this Maths.SE post) on $x\mapsto \frac{1}{x}\mathbb I_{\{x \neq 0\}}$ and what has already been established right above.
As for $\mathcal L^1$ norm, it is not even  clear if the sequences which converge to the real-valued $X, Y$ have their product $X_nY_n$ or $XY$ to be in $\mathcal L^1(\Omega). $ See this Maths.SE post for a counterexample.
For $\mathcal L^1$ convergence of sum of random variables, one can resort to the inequality (cf. $\rm [II]$)
$$ \mathbb E|(X_n -X) +(Y_n -Y) |^p\leq 2^p\left(\mathbb E|X_n- X|^p+\mathbb E|Y_n - Y|^p\right),~~p\in\mathbb R^{\geq 0}.\tag  4$$

Reference:
$\rm [I]$ Real Analysis: Theory of Measure and Integration, J.Yeh, World Scientific Publishing, $2014, $ chapter $1\S6, $ pp. $107-115,$ problems $6.9, ~6.13,~6.14.$
$\rm [II] $ Probability: A Graduate Course,  Allan Gut, Springer Science$+$Business, $2005, $ chapters $3, ~5,$  p. $127, ~247.$
