Two independent Poisson processes (no successive arrivals etc) 
Requests for orange juices and lemonades are according to two independent Poisson processes of rates $\lambda,~\mu$ (per minute), accordingly.
(1) What is the probability in the next one minute there are requests only for lemonades?
(2) Given that two requests were made in the last minute, what is the probability the times of request among them were done in less than 1/2 minute?
(3) What is the probability in the next minute no successive requests for the same type of drink are made?

Attempt Let $\{O_t\}, ~\{L_t\}$ the relative PP for oranges and lemonades, respectively.
(1) we get some ($\geq0$) lemonades and no orange juices in the next minute, so:
$$\sum_{k=0}^{\infty}P(L_1=k,~0_1=0)~P(L_1=k,~O_1=0/L_1+O_1=k)=
\sum_{k=0}^{\infty}e^{-(\lambda+\mu)}\frac{(\lambda+\mu)^k}{k!}\bigg(\frac{\mu}{\lambda+\mu}\bigg)^k=e^{-\lambda}.$$ 
(2) Let $S_1,~S_2$ be the (total) times of the first,  second arrival of $\{L_t+O_t\}$, respectively, so under our hypothesis they are uniformly distributed on $[0,1]$ and $$P(|S_2-S_1|<1/2)=1-2\times 1/8=3/4.$$ 
(3) This one i don't know how to approach.
Edit. An approach to (3). No successive arrivals for $O_1,~L_1$
means that we have $k$ total arrivals for some $k=0,1,2,\ldots$, with probability $e^{-(\lambda+\mu)}\frac{(\lambda+\mu)^k}{k!}$ and given the $k$ total arrivals, 
$\bullet$ for $k=2m$ even we have a sequence of arrivals $OLOL...OL$, with probability $\bigg(\frac{\lambda}{\lambda+\mu}\bigg)^m\bigg(\frac{\mu}{\lambda+\mu}\bigg)^m$ or a sequence of arrivals $LOLO...LO$, with the same probability,
$\bullet$ for $k=2m+1$ odd we have a sequence of arrivals $OLOL...OLO$, with probability $\bigg(\frac{\lambda}{\lambda+\mu}\bigg)^m\bigg(\frac{\mu}{\lambda+\mu}\bigg)^m\bigg(\frac{\lambda}{\lambda+\mu}\bigg)$ or a sequence of arrivals $LOLO...LOL$, with   probability $\bigg(\frac{\lambda}{\lambda+\mu}\bigg)^m\bigg(\frac{\mu}{\lambda+\mu}\bigg)^m\bigg(\frac{\mu}{\lambda+\mu}\bigg)$.
Working the sums, the desired probability is:
$$e^{-(\lambda+\mu)}\bigg(2\cosh(\sqrt{\lambda \mu})+\frac{1}{\sqrt{\lambda \mu}}\sinh(\sqrt{\lambda \mu})\bigg).$$
Thank you in advance.
 A: Here is a solution to part (1) that exploits the relationship between
Poisson and exponential disributions. Some of the rest might also benefit 
from this point of view.
Let $L$ be the waiting time for the first lemonade order.
Then $L \sim \mathsf{EXP}(rate = \mu = 2).$ That is, $F_L(t) = 1 - e^{-\mu t},$
for $t > 0.$  Similarly, let $J$ be the wait for the first orange juice order
with $F_J(t) = 1 - e^{-\lambda t},$ for $t > 0.$
Then 
$$P(L \le 1, J > 1) = P(L \le 1)P(J > 1) = F_L(1)(1-F_J(1)) \\= (1-e^{-\mu t})(e^{-\lambda t})
= e^{-\lambda t} - e^{-(\mu - \lambda)t}.$$
If $\mu = 2$ and $\lambda = 3$, then $P(L \le 1, J > 1) = 0.043.$

In the simulation below of a million realizations, rexp generates 
random exponential observations, pexp is the exponential CDF, exp is $e$ to a power, and
the mean of a logical vector (with T and F values) is its proportion of T's. The simulated  answer is  accurate to three places.
m = 10^6; a = rexp(m, 2);  j = rexp(m, 3)
mean(a < 1 & j > 1);  exp(-3)-exp(-5);  pexp(1,2)*(1-pexp(1,3))
## 0.042865     # aprx (1) by simulation
## 0.04304912   # exact (1)
## 0.04304912   # exact (1)

Note: If I understand what (2) is asking--and I'm not at all sure of 
that--then a rough simulated answer is about 0.82 (for my choice of parameters).
