It is not necessary to use any kind of bounds. The average number of fish you catch per day is $10$, so $\lambda=10$. The distribution of the number of salmon you catch per day is then $S\sim\text{Poisson}(10\times .3)=\text{Poisson}(3)$. The probability that $S\ge5=1-(.0498+.1494+.2240+.2240+.1680)=.1848$.
For part a, we could try $P(S=s \&T=t)$ (standing for salmon and trout caught) equals $\sum_{f=0}^\infty P(S=s\& T=t|F=f)P(F=f)=\sum_{f=0}^\infty P(S=s|T=t,F=f)P(T=t|F=f)P(F=f)$
Trying $P(S=0,T=0)=^? P(S=0)P(T=0)$ we get the rhs to be $e^{-.3\lambda}e^{-.4\lambda}=e^{-.7\lambda}$.
The lhs is found by the total law of probability to be $\sum_{f=0}^\infty P(S=0|T=0,F=f)P(T=0|F=f)P(F=f)=\sum_{f=0}^\infty (.5)^f(.6)^f\frac{e^{-\lambda}\lambda^f}{f!}=e^{-\lambda}\sum_{f=0}^\infty \frac{(.3\lambda)^f}{f!}$
Using the identity $e^\lambda=\sum_{x=0}^\infty \frac{\lambda^x}{x!}$, we get the preceding expression to be $e^{-\lambda}e^{.3\lambda}=e^{-.7\lambda}$ which, surprisingly or not, matches the marginals multiplied together. Now I dunno if this will hold if $P(S=1)P(T=1)=^?P(S=1,T=1)$. Edit: Holds as well. Now would hazard to surmise that holds for all fishes and all quantities; conclude independence.
Proof for salmon and trout (get ready...):
$$\begin{split}P(S=s \&T=t)&=\sum_{f=s+t}^\infty P(S=s\& T=t|F=f)P(F=f)\\
&=\sum_{f=s+t}^\infty P(S=s|T=t,F=f)P(T=t|F=f)P(F=f)\\
&=\sum_{f=s+t}^\infty \frac{e^{-\lambda}\lambda ^f}{f!}{f\choose t}.4^t.6^{f-t}{f-t\choose s}.5^{f-t}\\
&=e^{-\lambda}\sum_{f=s+t}^\infty \frac{\lambda ^ff!t!(f-t)!s!(f-t-s)!}{f!.4^t.3^{f-t}(f-t)!}\\
&=\frac{(.3\lambda)^s\lambda ^t e^{-\lambda}.4^t}{t!s!}\sum_{f=s+t}^\infty \frac{(.3\lambda)^{f-t-s}}{(f-t-s)!}\\
&=\frac{(.3\lambda)^s\lambda ^t e^{-\lambda}.4^t}{t!s!}\sum_{x=0}^\infty \frac{(.3\lambda)^{x}}{x!}\\
&=\frac{(.3\lambda)^s}{s!}e^{-.3\lambda}\frac{e^{-.4\lambda}(.4\lambda)^t}{t!}=P(S=s)P(T=t)\end{split}$$
The Markov bound is $P(S\ge5)\le\frac{E(S)}{5}=\frac 35=.6$, which is true as we found above.
The Chernoff bound is $P(S\ge 5)\le\min_{s>0}e^{-5s}e^{3(e^s-1)}$. Since $\log$ is increasing this is the same as minimizing the log of it, $-5s+3e^s-3$. The derivative wrt $s$ is $-5+3e^s=0$, with solution at $s=\log {\frac 53}$. The second derivative is positive, indicating a minimum. This gives the bound $e^{-5\log{\frac 53}+5-3}=0.574573$.
No, there is no overcounting, the numerator and denominator both use combinations, so they are consistent.
Let me illustrate with a simple example of drawing without replacement, and find the probability that $2$ red marbles and $3$ blue marbles are drawn from a total of $4$ red and $8$ blue marbles.
The combination method will give $\Large\frac{\binom42\binom83}{\binom{12}5}=\frac{14}{33}$
But if you try direct multiplicative way, you get
$\Large\frac4{12}\frac3{11}\frac8{10}\frac79\frac68 = \frac7{165}$ and you will need to multiply by $\binom52$ to get the right answer, because you counted the numerator in one particular order, while you counted the denominator in all possible orders.
Best Answer
Keeping it simple: