Poisson distribution is usually used to model number of accidents. Binomial distribution is not good here because it takes only values from 0 to N (you must specify N but it is not specified in the problem). Normally distributed random variable can take negative values. As the parameter lambda of Poisson distribution is equal to the mean of the random variable with this Poisson distribution then lambda=2 for a 3-month period. For a year period lambda=2*4=8 and for a month period lambda=2/3.
As we said before the parameter lambda of Poisson distribution was equal to the mean of the random variable with this Poisson distribution. The variance of this random variable is also equal to lambda. so for a year period mean=8 and standard deviation = SQR(variance)=2*SQR(2).
Probability of the event that no accident will happen this month is equal to the probability that a random variable with Poisson distribution (lambda=2/3 as it is a month period) is equal to 0. So it is EXP(-lambda)=EXP(-2/3)=0,5134. So approximately 51,34% of months are without accidents.
The probability of at least one accident for a month is 1 minus probability of the event that no accident happens this month. So it is equal to 1-0,5134=0,4866.