a. This is a first order linear PDE with constant coefficient.
\begin{equation}
\frac{d x}{4} = \frac{d y}{3} = \frac{d u}{0};
\end{equation}
Then we get $u(x, y) = f(3x+4y)$ for some arbitrary $f$.
b. Use IVP $u|_{x=0}=\sin (y)$, we get
\begin{equation}
f(4y) = \sin(y)\\f(w) = \sin(\frac{y}{4})
\end{equation}
Hence solution is
\begin{equation}
u(x,y) = f(3x+4y) = \sin(\frac{3}{4}x+y)
\end{equation}
c. With initial condition $u|_{x=0}=y$, $y>0$. Hence the solution is
\begin{equation}
u(x, y) = \frac{3}{4}x+y.
\end{equation}
Since $\forall x>0, y>0\implies \frac{3}{4}x+y >0$.
Thus this solution is defined on the whole domain $\{x>0,y>0\}$.
d. With initial condition $u|_{x=0}=y$, $y>0$. Hence the solution is
\begin{equation}
u(x, y) = \frac{3}{4}x+y.
\end{equation}
Since $f$ is define when $y>0$, the solution only is defined where $\frac{3}{4}x+y >0$, then when $y> -\frac{3}{4}x$, the solution is defined. However when $y< -\frac{3}{4}x$ we need to impose condition at $y = 0$, we get
\begin{equation}
f(3x) = x \\
f(w) = \frac{w}{3}\\
\end{equation}
Then we get:
\begin{equation}
u(x,y) = x + \frac {4}{3}y, (y< -\frac{3}{4}x)
\end{equation}
Final solution would be:
\begin{equation}
u(x,y) = \left\{
\begin{array}{l l}
\frac{3}{4}x+y & \quad y>-\frac{3}{4}x\\
x+\frac{3}{4}y & \quad y<-\frac{3}{4}x
\end{array} \right.
\end{equation}
