$1.$ Let the random variable $X$ be the demand. We are told that $X$ has Poisson distribution with mean (parameter) $\lambda=1.5$. Thus
$$\Pr(X=k)=e^{-\lambda} \frac{\lambda^k}{k!}\tag{$1$}$$
for every non-negative integer $k$.
(i) We want $\Pr(X=0)$. By $(1)$, this is simply $e^{-\lambda}$.
(ii) One has to interpret the meaning of "demand is refused." This seems to be an unusual way of saying that at least one customer is turned away, meaning that $X \ge 3$. To calculate $\Pr(X\ge 3)$, we find $\Pr(X=0)+\Pr(X=1)+\Pr(X=2)$, and subtract the sum from $1$.
We have $\Pr(X=0)=e^{-\lambda}$. By $(1)$, $\Pr(X=1)=\lambda e^{-\lambda}$ and $\Pr(X=2)=\frac{\lambda^2}{2!}e^{-\lambda}$. Add up. My calculator gives approximately $0.8088468$. Subtract this from $1$.
$2.$ Let the random variable $X_i$ be the number of bacteria in the $i$-th test tube ($i=1,2,\dots, 10$). The $X_i$, we are told, have Poisson distribution with mean (parameter) $\lambda=3$. We will assume that the $X_i$ are independent. This assumption really cannot be scientifically justified, and should have been mentioned explicitly in the statement of the problem.
For any $i$, $\Pr(X_i=0)=e^{-\lambda}$. So the probability that the $i$-th test tube shows some growth, meaning that $X_i \ge 1$, is $1-e^{-\lambda}$.
The probability that all the $X_i$ are $\ge 1$ is equal to $(1-e^{-\lambda})^{10}$. As a check on your calculations, the number turns out to be roughly $0.6000803$.
