If the letters of the word "about" are placed at random in a row, find the probability that three successive letters will be vowels.

Firstly, not quite sure how to set this up. Secondly, would this involve combinations and permutations? This is for an Algebra 1 class. Thanks!

There are $5!=120$ ways to arrange the word "about".

Group the $3$ vowels together, if you find the number of combinations with the vowel group and the two consonants the result will be $3!=6$, and there are $3!=6$ ways to group the $3$ vowels together so there will be $36$ ways that three successive letters will be vowels, so the answer is just $\frac{36}{120}=\frac3{10}$

Firstly, not quite sure how to set this up.

Count the arrangements of the symbols {a,b,o,u,t}

Count the arrangements of the symbols $\{aou, b, t\}$ that is, treating the three vowels as one symbol, count the arrangements when they are adjacent.

Count the arrangements of the symbols $\{a,o,u\}$, that is, count the ways the vowels can be arranged among themselves.

Put these counts together in some way to obtain the answer.

$$\dfrac{3!~3!}{5!}$$

Secondly, would this involve combinations and permutations?

Yes.

put the vowel sound word (a,o,u) in a group then the word arranging be (a, o, u) b and t , then the combination is 3! and the vowel sound word (a,o,u) can also arrange in 3! combination. So the possible arrangement is 3! x 3!= 6 x 6 = 36