I will use the following notation to be as consistent as possible with the wiki (in case you want to go back and forth between my answer and the wiki definitions for the poisson and exponential.)
$N_t$: the number of arrivals during time period $t$
$X_t$: the time it takes for one additional arrival to arrive assuming that someone arrived at time $t$
By definition, the following conditions are equivalent:
$ (X_t > x) \equiv (N_t = N_{t+x})$
The event on the left captures the event that no one has arrived in the time interval $[t,t+x]$ which implies that our count of the number of arrivals at time $t+x$ is identical to the count at time $t$ which is the event on the right.
By the complement rule, we also have:
$P(X_t \le x) = 1 - P(X_t > x)$
Using the equivalence of the two events that we described above, we can re-write the above as:
$P(X_t \le x) = 1 - P(N_{t+x} - N_t = 0)$
But,
$P(N_{t+x} - N_t = 0) = P(N_x = 0)$
Using the poisson pmf the above where $\lambda$ is the average number of arrivals per time unit and $x$ a quantity of time units, simplifies to:
$P(N_{t+x} - N_t = 0) = \frac{(\lambda x)^0}{0!}e^{-\lambda x}$
i.e.
$P(N_{t+x} - N_t = 0) = e^{-\lambda x}$
Substituting in our original eqn, we have:
$P(X_t \le x) = 1 - e^{-\lambda x}$
The above is the cdf of a exponential pdf.
Let $X$ and $Y$ denote the number of large claims and small claims respectively
in an year. Then, $N = X+Y$ is the total number of claims and is known to
be a Poisson$(\lambda)$ random variable. Now, given that $N=n$, the
conditions stated in the problem tell us that the conditional
distribution of $X$ is binomial with parameters $(n,p)$ and that of $Y$
is binomial with parameters $(n,1-p)$. Note that conditioned on $N=n$, $X$ and $Y$ are very much dependent random variables since $Y=n-X$.
But, unconditionally,
$X$ and $Y$ are independent Poisson$(\lambda p)$ and Poisson$(\lambda(1-p))$
random variables.
To see why all this is so, consider that for $0 \leq r \leq n$,
$$P\{X=r, Y=s \mid N=n\}
= \begin{cases} \displaystyle\binom{n}{r}p^r(1-p)^{s}, & \text{if} ~s = n-r,\\
0, & \text{if} ~ s \neq n-r.
\end{cases}$$
Consequently, for any $r, s \geq 0$,
$$\begin{align}
P\{X = r, Y = s\} &= \sum_{n=0}^\infty P\{X=r, Y=s, N=n\}\\
&= \sum_{n=0}^\infty P\{X=r, Y=s \mid N=n\}P\{N =n\}\\
&=\binom{r+s}{r}p^r(1-p)^{s}e^{-\lambda}\frac{\lambda^{r+s}}{(r+s)!}
&\scriptstyle{\text{only the $n=r+s$ term is nonzero in the sum}}\\
&= \frac{(r+s)!}{r!s!}(\lambda p)^r(\lambda(1-p))^{s}
e^{-\lambda p - \lambda(1-p)}\frac{1}{(r+s)!}\\
&= e^{-\lambda p}\frac{(\lambda p)^r}{r!}\cdot
e^{-\lambda(1-p)}\frac{(\lambda(1-p))^s}{s!}\\
&= P\{X=r\}P\{Y = s\}
\end{align}$$
showing that $X$ and $Y$ are independent Poisson$(\lambda p)$ and
Poisson$(\lambda(1-p))$ random variables respectively. Consequently, conditioned
on $X = r$, $Y$ continues to be a Poisson$(\lambda(1-p))$ random variable.
Best Answer
a) Yes, you're correct, the value of $\lambda$ should be 12.
What have you done for the other two?
b and c both look right to me.