Answer: (a) No; (b) No.

a. The integers $x$, $x+1$, $x+2$ are congruent to $0$, $1$, $2$ modulo $3$ in some order. The squares of these integers are congruent to $0$, $1$, $1$ modulo $3$, respectively. Hence the l.h.s. of the equation is congruent to $2$ while the r.h.s. is congruent to $0$ or $1$ modulo $3$. Thus the equality cannot hold.

b. Among the integers $x$, $x+1$, $x+2$, $x+3$, there are two even numbers and two odd numbers. The squares of even numbers are divisible by $4$ while the squares of odd numbers are congruent to $1$ modulo $4$. Thus the l.h.s. of the equation is congruent to $2$ while the r.h.s. is congruent to $0$ or $1$ modulo $4$. Hence the equality cannot hold.