How do you get $d$ in textbook RSA So people say to choose $d$ such that $e*d=1(mod((p-1)*(q-1)))$.

However, $1\%$ anything always equals $1$. Maybe I am not understanding it correctly, but the only way to get $e \cdot d=1$ is just by doing $e^{-1}$ which leads into a very very small number $\lll 1$ because $e$ is large.

How are people getting $e$ and $d$ such that they are both large numbers?

My main confusion really just comes from that $1\%$ anything always equals $1$.
There are no percentages, fractions, real or floating point numbers involved. 

When using modular arithmetic, the "division" operation is defined differently. If you have some values $a, b$ that have been multiplied together into $ab$, what you want from division is that $\frac{ab}{b} = a$, or stated alternatively that $\frac{a}{a} = 1$.

When working with modular arithmetic, you need a different division operation than the "normal" one taught in basic math classes. This is easy to see with an example. Suppose that you have $$a=5\\b=7\\N=19\\ab \equiv 16 \bmod N$$ It is easy to see that dividing (in the traditional sense) $16$ by $7$ is not going to yield $5$.

Instead, you can multiply $16$ by $11$, which happens to be equivalent to $5$ modulo $19$. This fulfills the behavior expected of the division operation, which is what we really care about.

A value $\frac{1}{b}$ such that $\frac{ab}{b} \equiv a \bmod N$ is called a modular multiplicative inverse.

Back to the specific question

In the case of RSA, $e$ and $d$ are modular multiplicative inverses: $ed \equiv 1 \bmod \lambda(N)$. 

Alternatively, you can denote $d$ by $\frac{1}{e}$ or $e^{-1}$; These are different ways of writing the same value.

A multiplicative inverse exists if the element is coprime to the modulus $N$. To actually find a modular multiplicative inverse of an element, you can use the extended euclidean algorithm.