Answer:
a)
So with the p value obtained and using the significance level given
we have
so we can conclude that we have enough evidence to reject the null hypothesis.
b) 

Step-by-step explanation:
Notation and hypothesis
When we have two independent samples from two normal distributions with equal variances we are assuming that
And the statistic is given by this formula:
Where t follows a t distribution with
degrees of freedom and the pooled variance
is given by this formula:
This last one is an unbiased estimator of the common variance
Part a
The system of hypothesis on this case are:
Null hypothesis:
Alternative hypothesis:
Or equivalently:
Null hypothesis:
Alternative hypothesis:
Our notation on this case :
represent the sample size for group 1
represent the sample size for group 2
represent the sample mean for the group 1
represent the sample mean for the group 2
represent the sample standard deviation for group 1
represent the sample standard deviation for group 2
First we can begin finding the pooled variance:
And the deviation would be just the square root of the variance:
Calculate the statistic
And now we can calculate the statistic:
Now we can calculate the degrees of freedom given by:
Calculate the p value
And now we can calculate the p value using the altenative hypothesis:
Conclusion
So with the p value obtained and using the significance level given
we have
so we can conclude that we have enough evidence to reject the null hypothesis.
Part b
For this case the confidence interval is given by:

For the 99% of confidence we have
and
and the critical value with 25 degrees of freedom on the t distribution is 
And replacing we got:

