$$x(x+3)=154\implies x^2+3x-154=0\implies(x+14)(x-11)=0\implies x=-14,11$$Both of these values give $(x+1)(x+2)=\boxed{156}$

\begin{align*}x(x+3)&=154\\x^2+3x&=154\\x^2+3x+2&=156\\(x+1)(x+2)&=\boxed{156}\end{align*}

This is just a binomial! $x(x+3) = 154$, so $x^2 + 3x -154 = 0$ Factor this to find the real value for x, and then plug it in to the other equation! Or, you can see that $154$ is $14*11$ . $14 = 11 + 3$, so $x=11$. Remember, when you can't do something smart, do something stupid. Either way, $12 \cdot 13 = 156$