$\frac{7}{5} \equiv 11 \pmod{12}$. Why is it $11$? I know this equation is true but I don't get the reason $\frac{7}{5}$ is congruent to $11$ here.
The quotient is supposed to be $1.4$ in non modular arithmetic and I don't get where $11$ has come from.
 A: The multiplication modulo $12$ is rather different from that in the rational numbers. But the understanding of division is the same. We want to be able to say (working modulo $12$) that if $$a\cdot b=c$$ then $$b=\frac ca$$
And that is what is happening here, because we have $5\cdot 11\equiv 7 \bmod 12$ and the fraction is just another way of expressing the same fact. You don't have $5\cdot 11=7$ in the integers or rationals.

As an aside, there is an additional issue here. We know that in the integers we can't divide by zero. Modulo $12$ we have "zero divisors" as well - for example $2\cdot 6 \equiv 3\cdot 4 \equiv 0$. We are always safe if we are dividing by numbers which are coprime to $12$, but multiples of $2$ or $3$ need special treatment - either there are no answers, or there are multiple possibilities. For example $$2\times 3\equiv 6 \times 3\equiv 10\times 3\equiv 6$$ so there is no definite value for $\frac 63$ which could be equal to $2, 6, 10$ by analogy with the usual division.
A: $$5 \cdot 11 = 55 \equiv 7 \mod 12.$$
In general, by $\frac{a}{b}$ we mean a number that, when multiplied by $b$, will give the value $a$. This definition is also used in modular arithmetic. So $\frac{7}{5} \equiv 11 \mod 12$, because both give the value $7$ when multiplied by $5$ (modulo $12$).
A: You can't just divide by $5$ and look at $1.4$. Instead you have to write $7/5=7\cdot 5^{-1}\mod 12$ and thus have to find the inverse of $5\!\!\!\mod \!\!12$. As $5\cdot 5=25\equiv 1\mod 12$ we have: $$7\cdot 5^{-1}=7\cdot 5=35\equiv 11 \mod 12.$$
A: $\frac{7}{5}$ is not an integer, the congruence relation $\bmod n$ is defined over the integers only.
On the other hand there is such thing as an inverse when working with modular arithmetic. Just like when working in the real numbers we have an inverse for $a$, we also have inverses in modular arithmetic. The inverse of the integer $a\bmod n$ is an integer $b$ so that $ab\equiv 1 \bmod n$ (just like in real numbers $a\cdot\frac{1}{a}=1$), we shall denote the inverse of $a$ as $a^{-1}$. It is also interesting to note not every integer has an inverse $\bmod n$ , in general $a$ has an inverse $\bmod n$ if and only if $a$ and $n$ are relatively prime.
So what is the inverse of $5\bmod 12$? it is the number $k$ so that $ak\equiv 1\bmod 12$. We can find it by trying:
$1\cdot 5=5$ no 
$2\cdot 5=10$ no 
$3\cdot 5=3$ no 
$4\cdot 5=8$ no 
$5\cdot 5=1$ yes
$6\cdot 5=6$ no 
$7\cdot 5=11$ no 
$8\cdot 5=4$ no 
$9\cdot 5=9$ no 
$10\cdot 5=2$ no 
$11\cdot 5=7$ no 
Hence the inverse of $5$ is $5$ itself (this rarely happens).
With this in mind we can know calculate $7\cdot5^{-1}=7\cdot5=35\equiv 11\bmod 12$

There are much better ways to obtain the inverse of a number, such as logarithmic exponentiation along with fermat or euler's theorem or via the euclidean algorithm.
A: The other answers have pointed out the right way to make the calculation. You should refer to them as they promptly answer your question. But I'll try to focus on the implicit question... the "$1.4$" confusion.
First, what is $1.4$? This is a representation. Namely, the decimal representation. That's how you should think of it: a name. It is like how $2$  is equal to  $10$ in binary. The "number" $1.4$, in reality, is $14.10^{-1}$. 
The decimal representation comes from a fact in analysis that we can always write a given real number $x$ as a series $x=\sum c_n10^n$. Then, we represent $x$ as $c_nc_{n-1}...c_0.c_{-1}c_{-2}...$.
So, your question has in fact a chance to be interpreted in this context: Is $7.5^{-1}=1.4$? Namely: is $7.5^{-1}=1+4.10^{-1}$?. It turns out that your question, sadly, does not make sense: $10$ is not invertible in your ring!
This goes on to telling: carrying on a representation that has in itself so much "analysis" in its heart as the decimal representation (note that real numbers will be series: limits.) has to be taken with great care.
Now, suppose you were at $\mathbb{Z}_{11}$
Does the equation $7.5^{-1}=1+4.10^{-1}$ make sense? Yes! $10^{-1}=-1$ (and indeed, equal to $10$ itself).
So, let's check if the equality holds: 
On one side: $7.5^{-1}=7.(-2)=-14=-3$
On the other side: $1+4.10^{-1}=1-4=-3$
So, in fact, we can say that $7.5^{-1}$ equals $1.4$ in $\mathbb{Z}_{11}$! But note that we have to explicitly expose what we mean by $1.4$, and that things don't carry on as easily. Note also that $1.4$ is simply a name for $-3$ here.
A: $$
5 \cdot 11 = 55 = \underbrace{12 \cdot 4}_{\equiv 0} + 7 \equiv 7 \mod 12.
$$
