The answer to your second question will help illuminate a different way of thinking about the first.
If the $5$ people were to sit in a straight line, then the number of ways they can sit is indeed $5!$. Since we are asking them to sit in a circle, each possible circular arrangement corresponds to $5$ different linear arrangements, depending on which seat we label the 'first' seat. This shows that there are $5!/5=4!=24$ ways for $5$ people to sit around a table.
Now, of all $24$ possible seating arrangments, only $2$ of them are arranged in increasing or decreasing order of age, so the desired probability is $1/12$.
Notice our answers are the same, but it is important to see the distinction in our reasoning. In a sense, you've chosen a first seat, which increased your numerator and denominator by a factor of $5$. Since the question asks for a ratio, this didn't affect the final answer, but we should still note the difference between sitting around a table, and sitting in a line.
We assign seats at random, by putting place mats labelled $1,2,3,\dots, A, B, C$ at random in front of the $10$ chairs. The place mats labelled $A,B,C$ indicate where any person apart from $1$ to $7$ may sit. A pair of lovers now enters the room, and they wish to sit together. We find the probability that they can do so without asking anybody to move.
The only thing that matters is the placement of $A, B, C$. And since the table is round, all that matters is the placement of $B$ and $C$ relative to $A$. So we have $9$ empty spaces, and have to choose $2$ of them. There are $\binom{9}{2}$ choices, all equally likely.
We count the bad choices. For a choice to be bad, it must involve choosing $2$ non-adjacent seats from the $7$ not next to $A$. There are $\binom{7}{2}$ ways to choose $2$ seats, of which $6$ give an adjacent pair. So there are $15$ bad choices, out of the $\binom{9}{2}$ choices. Thus the probability there will be an adjacent pair is $\frac{21}{36}$.
Best Answer
If no-one can sit next to someone else, then three seats are blocked every time someone sits down. However, some of these may have been blocked already, by previous sitters. So, the first person to sit down has $15$ choices, and leaves twelve choices. The next person to sit down has $12$ choices; in two of these cases, he leaves ten choices; in the other ten cases, he leaves nine. The final person to sit down has either $10$ or $9$ choices, depending on what the second person did. The result is $$ N=15\cdot(2\cdot10+10\cdot9)=1650 $$ ways for three (distinguishable) people to sit down. The probability, then, is $$ p=\frac{N}{15\cdot14\cdot13}=\frac{1650}{2730}=\frac{55}{91}. $$