The given problem:
Use Lemma 2.3.3 together with Fermat's little theorem to show that 91 is a pseudoprime to the base 3.
Lemma 2.3.3. Let $m_1 \dots m_r \in $ N. If $a \equiv b \pmod {m_i}$, $\forall i =1, \dots, r$, then $a \equiv b \pmod {{\rm lcm}(m_1, \dots, m_r)}$. (If $\gcd(m_i, m_j)=1$ when $i \neq j$, so $a \equiv b \pmod {m_1 \dots m_r}$.)
Theorem 2.4.5 (Fermat's little theorem II). Let $p$ prime and $a \in$ Z such that $p \nmid a$. Then $a^{p-1} \equiv 1 \pmod p$.
My own attempt: Because $91=7 \cdot 13$, the number is composite. According to Theorem 2.4.5 $3^6 \equiv 1 \pmod 7$. On the other hand $90 = 6 \cdot 15 $, so $3^{90} = 3^{6 \cdot 15} = (3^6)^{15} \equiv 1^{15} \equiv 1 \pmod 7$ Then let's assume according to Theorem 2.4.5 that $3^{12} \equiv 1 \pmod {13}$. On the other hand $90=12 \cdot 7+6$. So $3^{90}=3^{12\cdot7+6}=3^{12\cdot7}\cdot 3^6 = (3^{12})^7\cdot3^6 \equiv 1^7 \cdot 3^6 \equiv 1 \pmod {13}$. So according to Lemma 2.3.3 we have $3^{90} \equiv 1 \pmod {91}$ so $3^{91} \equiv 3 \pmod {91}$. So 91 would be a pseudoprime to the base 3.