Answer:
a) 
Using the symmetrical property we can write this like this:

We can solve for the probability like this:


And we can find the value using the following excel code: "=T.INV(0.05,7)"
So on this case the answer would be 
b) For this case we can use the complement rule and we got:

We can solve for the probability and we got:

And we can use the following excel code to find the value"=T.INV(0.95;7)"
And the answer would be 
Step-by-step explanation:
Previous concepts
The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".
The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.
The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."
Solution to the problem
For this case we know that 
And we want to find:
Part a

We can rewrite the expression using properties for the absolute value like this:

Using the symmetrical property we can write this like this:

We can solve for the probability like this:


And we can find the value using the following excel code: "=T.INV(0.05,7)"
So on this case the answer would be 
Part b

For this case we can use the complement rule and we got:

We can solve for the probability and we got:

And we can use the following excel code to find the value"=T.INV(0.95;7)"
And the answer would be 