On the incoming, $x<0$, side of the mass there is the incident wave and the reflected wave so $$y=e^{i\omega(t-x/c)}+Ae^{i\omega(t+x/c)}$$ where the incident wave has an amplitude of unity as described in the problem statement and the reflected wave has an unknown amplitude, $A$, and is traveling in the opposite direction. On the outgoing, $x>0$, side of the mass there is only the transmitted wave so $$y=Be^{i\omega(t-x/c)}\text{.}$$
The string must me continuous at $x=0$ so $A=B-1$. Continuity also implies that there can't be a phase change other than 0 or $\pi$ at $x=0$, which is why I haven't bothered to include phases in the three terms of $y$.
The boundary condition at $x=0$ (note that it doesn't matter which whether we pick the $x<0$ or $x>0$ version of $y$ for the left hand side, we'll get the same answer) gives
$$ -BM\omega^2e^{it\omega}=\frac{iT\omega}{c}e^{it\omega}(1-A-B) $$
Applying continuity makes this
$$ -BM\omega^2e^{it\omega}=\frac{2iT\omega}{c}e^{it\omega}(1-B)\text{.} $$
Solving for $B$ gives
$$B=\frac{2T}{2T+icM\omega}=\frac{4T^2}{4T^2-c^2M^2\omega^2}-\frac{2iTcM\omega}{4T^2-c^2M^2\omega^2}$$
which in turn gives $$A=\frac{-icM\omega}{2T+icM\omega}=\frac{c^2M^2\omega^2}{4T^2-c^2M^2\omega^2}-\frac{2iTcM\omega}{4T^2-c^2M^2\omega^2}\text{.}$$
Strictly speaking, we now know everything about the reflected and transmitted waves. To visualize the motion of the string, we can take the real part of our solution to the wave equation.
For the $x<0$ side this gives
$$ y=\cos[\omega(t-x/c)]+\frac{cM\omega}{4T^2-c^2M^2\omega^2}\left ( cM\omega\cos[\omega(t-x/c)]+2T\sin[\omega(t-x/c)] \right ) $$
For the $x>0$ side we have
$$ y=\frac{2T}{4T^2-c^2M^2\omega^2}\left ( 2Tcos[\omega(t-x/c)]+cM\omega\sin[\omega(t-x/c)] \right ) $$ 402