Answer
a) $12.9\;\rm N$
b) $2.0\;\rm cm$
c) $12.76\;\rm m/s$
Work Step by Step
We are given a function of a wave
$$D_{(x,t)}=A\sin\left[ \frac{2\pi}{\lambda}x-2\pi f t+\phi_0 \right]=2\sin\left[ 12.57x-638t+0\right]$$
Thus, $k=12.57\;\rm rad/m$, $\omega=638\;\rm rad/s$, and $\phi_0=0\;\rm rad$
a) We know that the tension in the string is given by
$$v=\sqrt{\dfrac{T_S}{\mu}}$$
where $T_S$ is the tension in the string, $\mu$ is its linear density, and $v$ is the speed of the wave in a stretched string.
Thus,
$$T_s=\mu v^2$$
where $v=\lambda f$, whereas $\dfrac{2\pi}{12.57}={\lambda} $, and $f=\dfrac{638}{2\pi}$
$$T_s=\mu \lambda^2f^2$$
Plugging the known;
$$T_s=(5\times 10^{-3})\left(\dfrac{2\pi}{12.57}\right)^2\left(\dfrac{638}{2\pi}\right)^2=\color{red}{\bf 12.88}\;\rm N $$
b) The maximum displacement of a point on the string is given by
$$D_{max}=A=\color{red}{\bf 2.0}\;\rm cm$$
c) The maximum speed of a point on the string is given by
$$v_{max}=A\omega=(2\times 10^{-2})(638)=\color{red}{\bf 12.76}\;\rm m/s$$
