a) Set $S = \frac{1}{a+1} + \frac{1}{a+2} + \frac{1}{a+3}$ and notice that $\frac{3}{a+3} < S < \frac{3}{a+1}$ to infer from $\frac{1}{4} < S < \frac{1}{3}$ that $\frac{1}{4} < \frac{3}{a+1}$ and $\frac{3}{a+3} < \frac{1}{3}$. Consequently $6 < a < 11$, so $a \in \{7, 8, 9, 10\}$. It is easy to check that $a = 7$ fails and $8, 9, 10$ are solutions.

b) Select $a_1 = p^2 + 1, a_2 = p^2 + 2, \dots, a_p = p^2 + p$ to get
$$S = \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_p} = \frac{1}{p^2+1} + \frac{1}{p^2+2} + \dots + \frac{1}{p^2+p}$$
and notice that
$$\frac{1}{p^2+1} > \frac{1}{p^2+2} > \dots > \frac{1}{p^2+p}$$
implies
$$\frac{p}{p^2+p} < S < \frac{p}{p^2+1} < \frac{p}{p^2} = \frac{1}{p}.$$