FA for w when length of w is even or divisible by 3
A simpler solution is to use a DFA with 6 states which count the number of letters seen so far, modulo 6. This is a DFA with 6 states $q_0,\ldots,q_5$, and transition function $\delta(q_i,\sigma) = q_{i+1 \bmod{6}}$. The accepting states are $\{q_0,q_2,q_3,q_4\}$, since a number is even or divisible by 3 iff its residue modulo 6 is one of 0,2,3,4.

Using Myhill–Nerode theory, you can easily show that this is the minimal DFA for the language. Using an NFA with multiple initial states, you can improve this to 5 states: have two starting states $q_0^{(2)},q_0^{(3)}$ which also double as accepting states, states $q_1^{(2)},q_1^{(3)},q_2^{(3)}$, and transitions $\delta(q_i^{(m)},\sigma) = \{q^{(m)}_{i+1 \bmod{m}}\}$. This is likely the minimal number of states for this type of NFA.