Answer
See the detailed answer below.
Work Step by Step
a) The wave is moving in the $x$-direction since it is $D_{(x,t)}$. And it is moving in the negative direction of the $x$-axis since the sign in front of $t$ is positive.
b) Knowing that
$$D_{(x,t)}=A\sin\left( kx-\omega t+\phi_0\right)=A\sin\left( \frac{2\pi}{\lambda}x-\frac{2\pi }{T} t+\phi_0\right)$$
Hence,
$$A\sin\left( \frac{2\pi}{\lambda}x-2\pi ft+\phi_0\right)=3.0\times 10^{-2}\sin\left( \frac{2\pi}{2.4}x+ \frac{2\pi }{0.2}t+2\pi \right)$$
Thus the frequency is given by
$$\frac{2\pi}{0.2}=2\pi f$$
$$f= \color{red}{\bf 5}\;\rm Hz$$
The wavelength
$$\dfrac{2\pi}{\lambda}=\dfrac{2\pi}{2.4}$$
so
$$\lambda =\color
{blue}{\bf 2.4}\;\rm m$$
The wave speed is given by
$$v=\lambda f=(2.4)(5)=\color
{red}{\bf 12}\;\rm m/s$$
The wave number
$$k=\dfrac{2\pi}{\lambda}=\dfrac{2\pi}{2.4} =\color
{red}{\bf 2.62}\;\rm rad/m$$
c)
$$D_{(0.2,0.5)}=3.0 \sin\left( \frac{2\pi}{2.4}(0.2)+ \frac{2\pi }{0.2}(0.5)+2\pi\right)=\color{red}{\bf -1.5}\;\rm cm$$