Let A = {1, 2, 3, 4, 5} and B = {a, b, c}.

(a) Each element of $A$ has 3 choices: $3^5 = 243$.

(b) Functions that miss a given element of $B$: $2^5$; that miss two given elements: $1$. By inclusion and exclusion, onto functions number $3^5 - \binom31 2^5 + \binom32 1^5 = 243 - 96 + 3 = 150$.

(c) From $B$ to $A$, the three elements take different images: $5 \times 4 \times 3 = 60$.

(d) Choose the two elements of the range in $\binom32 = 3$ ways; the functions onto those two number $2^5 - 2 = 30$. So $3 \times 30 = 90$. (Check: $150 + 90 + 3 = 243$, with 3 constant functions.)

(e) Such a function is fixed once we know how many elements go to 1, to 2 and to 3: whole numbers $n_1 + n_2 + n_3 = 5$. By stars and bars there are $\binom{5 + 2}{2} = 21$.