For part (a) we should choose $x, y$ such that $y+1 | x$ and $y | x+1$. Then $ {\rm lcm}(x,y+1)\cdot{\rm lcm}(x+1,y) = x \cdot (x+1)$ clearly. But by CRT the system $x \equiv 0 \pmod{y+1}$ and $x \equiv -1 \pmod{y}$ has infinitely many solutions for $x$ because $\gcd(y, y+1)$. So we find infinitely many $x$. (b) Once again, it would be nice for $\text{lcm}(x, y+1) = y+1$ and $\text{lcm}(x+1, y) = y$. So we would need that $x | y+1$ and $x+1 | y$ (*). So we should just construct optimal $y$. The choice $y = 2^{2^{9001}}-1$ is a good choice. Then any $x = 2^{2^k}$ for $k = 0, 1, \ldots, 9000$ satisfies (*) This finishes the construction.

Take x = (y+1)(ky-1) for a.