What's the probability that you wait $2$ hours for the train when you have already waited $1$ hour? 
Let's say there are independent waiting times of trains which are
  exponentially distributed with mean value $\frac{1}{2}$ hours. If you
  have already waited $1$ hour for the train, what's the probability
  that you will wait $2$ hours?

We have that mean value $\mu = \frac{1}{2}$. It's known that $\mu = \frac{1}{\lambda} \Leftrightarrow \frac{1}{2} = \frac{1}{\lambda} \Leftrightarrow \lambda = 2$
Now comes the part I'm not sure about.. when you have already waited $1$ hour before and now wait additional $2$ hours how the probability is calculated correctly here? I have looked this up on the internet and found that memorylessness applies to exponential distribution with formula:
$$P(X > r+t \mid X > r) = P(X>t)$$
where $r$ is the time you have previously waited, so $r=1$ and where $t$ is the time you have waited afterwards, thus $t=2$. Putting this into the formula we have
$$P(X>3 \mid X>1) = P(X>2) = 1-P(X<2) = 1-\left(1-e^{-2 \cdot2}\right) \approx 0.0183$$
So when you have already waited $1$ hour for the train, there is a probability of $0.0183$ that you will be waiting additional $2$ hours for the train?
Can you please tell me if it's correct like that because that's how I would do it in the exam :c

Let $T_n\stackrel{\mathrm{i.i.d.}}\sim\mathrm{Expo}(\lambda)$. Define $S_0=0$ and $S_n=\sum_{i=1}^n T_n$. Then $S_n$ is a renewal process, in fact a Poisson process, with associated counting process $N(t) = \sup\{n: S_n\leqslant t\}$. Define the age process by $A_t = t - S_{N(t)}$ and the residual process by $R_t = S_{N(t)+1} - t$ for $t\geqslant 0$. Define the renewal function $M(t) = \mathbb E[N(t)]$. Since $N(t)$ is Poisson distributed with mean $\lambda t$, it readily follows that $M(t)=\lambda t$. Let $Z_t\stackrel{\mathrm{def}}=(A_t,R_t)$. We derive the distribution of $Z_t$ by considering when $A_t=t$, that is, when $t$ lies within the first renewal interval. Then we have
$$
f_{Z_t}(t,y) = f_{T_1}(t+y) = \lambda e^{-\lambda(t+y)}\cdot\mathsf 1_{[0,\infty)}(y).
$$
The other case where $A_t<t$ corresponds to when $t$ occurs after the first renewal. We have
\begin{align}
f_{Z_t}(x,y) &= \sum_{n=1}^\infty \frac{\lambda^n(t-x)^{n-1}e^{-\lambda(t-x)}}{(n-1)!}\lambda e^{-\lambda(x+y)}\
&= \lambda^2 e^{-\lambda(x+y)}\cdot\mathsf 1_{[0,t)\times[0,\infty)}(x,y).
Combining these two cases yields
$$
f_{Z_t}(x,y) = \lambda^2e^{-\lambda(x+y)}\mathsf 1_{[0,t)\times[0,\infty)}(x,y) + \lambda^2e^{-\lambda(t+y)}\mathsf 1_{\{x=t\}\times[0,\infty)}(x,y)
$$
