I’ll do a very careful job of proving the first one; see if that’s enough for you to go on and do the others on your own.
To show that $\dfrac{x_n}{y_n}\to\dfrac{a}b$, we must show that for every $\epsilon>0$ there is an $N_\epsilon\in\Bbb N$ such that $\left|\frac{x_n}{y_n}-\frac{a}b\right|<\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;,$ so start by fixing $\epsilon>0$.
We know that by taking $n$ large enough we can get $x_n$ very close to $a$ and $y_n$ very close to $b$; how close do we need them to get in order to ensure that $\left|\frac{x_n}{y_n}-\frac{a}b\right|<\epsilon\;?$
A little algebra is in order: $\left|\frac{x_n}{y_n}-\frac{a}b\right|=\left|\frac{bx_n-ay_n}{by_n}\right|=\frac{|bx_n-ay_n|}{|by_n|}\;.$ Now $b\ne 0$, so there is an $M\in\Bbb N$ such that $|y_n|>\frac12|b|>0$ for all $n\ge M$, and for all $n\ge M$ we then have $\frac{|bx_n-ay_n|}{|by_n|}<\frac{|bx_n-ay_n|}{\frac12b^2}=\frac2{b^2}|bx_n-ay_n|\;.$ If we can find an $M_\epsilon\in\Bbb N$ such that $|bx_n-ay_n|<\frac{b^2}2\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;,$ we’re done: we let $N_\epsilon=\max\{M,M_\epsilon\}$, and for each $n\ge N_\epsilon$ we have
$\left|\frac{x_n}{y_n}-\frac{a}b\right|=\frac{|bx_n-ay_n|}{|by_n|}<\frac2{b^2}|bx_n-ay_n|<\frac2{b^2}\cdot\frac{b^2}2\epsilon=\epsilon\;,$ as desired.
Now we use the trick of subtracting and adding the same thing and apply the triangle inequality:
$\begin{align*} |bx_n-ay_n|&=|bx_n-ba+ab-ay_n|\\ &\le|bx_n-ba|+|ab-ay_n|\\ &=|b||x_n-a|+|a||b-y_n| \end{align*}$
Since $\langle x_n:n\in\Bbb N\rangle$ converges to $a$, we know that there is a $K_1\in\Bbb N$ such that $|x_n-a|<\dfrac{|b|\epsilon}4$ whenever $n\ge K_1$, and this ensures that $|b||x_n-a|<\dfrac{b^2\epsilon}4$ whenever $n\ge K_1$. If $a=0$, that already ensures that $|bx_n-ay_n|\le|b||x_n-a|+|a||b-y_n|=|b||x_n|<\frac{b^2\epsilon}4<\frac{b^2\epsilon}2\;,$ and we don’t have to worry about the $|a||b-y_n|$ term. If $a\ne 0$, the fact that $\langle y_n:n\in\Bbb N\rangle$ converges to $b$ ensures that there is a $K_2\in\Bbb N$ such that $|y_n-b|<\dfrac{b^2\epsilon}{4|a|}$ for all $n\ge K_2$. Thus, $|bx_n-ay_n|<|b||x_n-a|+|a||b-y_n|<\frac{b^2\epsilon}4+|a|\frac{b^2\epsilon}{4|a|}=\frac{b^2\epsilon}2$ whenever $n\ge\max\{K_1,K_2\}$.
If $a=0$ let $M_\epsilon=K_1$, and if $a\ne 0$ let $M_\epsilon=\max\{K_1,K_2\}$; then for all $n\ge K$ we have $|bx_n-ay_n|<\frac{b^2\epsilon}2\;,$ which is exactly what we needed: with $N_\epsilon=\max\{M,M_\epsilon\}$ we now know that $\left|\frac{x_n}{y_n}-\frac{a}b\right|<\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;,$ which is exactly what it means to say that $\left\langle\frac{x_n}{y_n}:n\in\Bbb N\right\rangle$ converges to $\dfrac{a}b$.
Added: I’ll do part (b) of (3). The hypotheses are that $b_n<0$ for each $n\in\Bbb N$ and that $\lim\limits_{n\to\infty}b_n=0$, and we want to show that $\lim\limits_{n\to\infty}\frac1{b_n}=-\infty$. The first step is to write down exactly what this means: we’re trying to show that for each $C<0$ there is an $N_C$ such that $\frac1{b_n} whenever $n\ge N_C$, so start by fixing some $C<0$.
Now look at what you have to work with: $\lim\limits_{n\to\infty}b_n=0$, so for each $\epsilon>0$ there is an $N_\epsilon\in\Bbb N$ such that $|b_n-0|<\epsilon$ whenever $n\ge N_\epsilon$. You also know that each $b_n$ is negative, so you can rewrite this as follows:
for each $\epsilon>0$ there is an $N_\epsilon\in\Bbb N$ such that $-\epsilon whenever $n\ge N_\epsilon$.
If $-\epsilon, then $0<-b_n<\epsilon$, so $\dfrac1{-b_n}>\dfrac1\epsilon$, and $\dfrac1{b_n}<-\dfrac1\epsilon$. Thus, for any $\epsilon>0$ we know that $\frac1{b_n}<-\frac1\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;.\tag{1}$ We’re trying to fine an $N_C$ such that $\frac1{b_n} $(1)$ and $(2)$ look very similar. In fact, if we let $\epsilon=-\frac1C$ and then set $N_C=N_\epsilon$, they become identical.
So that’s what we do: given $C<0$, let $\epsilon=-\frac1C$. By hypothesis there is an $N_\epsilon\in\Bbb N$ such that $(1)$ is true. Now let $N_C=N_\epsilon$, and we see that $(2)$ is true: we’ve found the needed $N_C$. This is exactly what’s required in order to conclude that $\lim\limits_{n\to\infty}\frac1{b_n}=-\infty$.
Added2: For part (a) of (3) we’re assuming that $\lim\limits_{n\to\infty}a_n=-\infty$, where $a_n\ne 0$ for $n\in\Bbb N$, and we want to prove that $\lim\limits_{n\to\infty}\frac1{a_n}=0$. In order to do this, we must show that for each $\epsilon>0$ there is an $N_\epsilon\in\Bbb N$ such that $\left|\frac1{a_n}-0\right|<\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;.\tag{3}$ We can simplify our goal a little by rewriting $(3)$: we want to show that for each $\epsilon>0$ there is an $N_\epsilon\in\Bbb N$ such that $\frac1{|a_n|}<\epsilon\quad\text{whenever}\quad n\ge N_\epsilon\;.\tag{4}$
What do we have to work with? We know that $\lim\limits_{n\to\infty}a_n=-\infty$, which by definition means that for each $C<0$ there is an $N_C\in\Bbb N$ such that
$a_n Now stop and think about what $(4)$ and $(5)$ say. $(5)$ says that we can make $a_n$ a negative number of large absolute value by taking $n$ large enough; that’s what we know that we can do. $(4)$, says that we can make $\frac1{|a_n|}$ very small by taking $n$ large enough; that’s what we want to be able to do. But making $a_n$ a negative number of large magnitude automatically makes $\frac1{|a_n|}$ a very small number, so if we can do the one, we can do the other. All we have to do is write down the details.
Let $\epsilon>0$ be any positive real number, and let $C=-\frac1\epsilon$. Then $C<0$, so by hypothesis there is an $N_C\in\Bbb N$ such that $(5)$ holds. Let $N_\epsilon=N_C$. Then for each $n\ge N_\epsilon$ we know that$n\ge N_C$, so $a_n Multiply $(6)$ by $-1$: $-a_n>-C=\frac1\epsilon\;.$ These are positive numbers, so taking reciprocals yields $-\dfrac1{a_n}<\epsilon$ for each $n\ge N_\epsilon$. Finally, $a_n$ is negative, so $-a_n=|a_n|$, and we’ve just shown $(4)$: that $\dfrac1{|a_n|}<\epsilon$ whenever $n\ge N_\epsilon$. This is precisely what is meant by the statement that $\lim\limits_{n\to\infty}\frac1{a_n}=-\infty$, so we’re done.
All of these arguments have worked the same way. I’ve used the definition of limit to write down exactly what I need to show and exactly what information I’m given. I’ve then tried to see how the two are related. In (3) it’s basically just been a matter of recognizing that when $|x|$ gets large, $\frac1x$ gets close to $0$, and vice versa. In (1) I actually had to work quite a bit harder, because the connection between what I was given and what I wanted was much less obvious. That’s where a certain amount of experience and practice are helpful: that trick of subtracting and adding the same intermediate quantity and then using the triangle inequality is very common, but it takes some ingenuity to discover it if you’ve never seen it before.