Answer
$\textbf{(a)}$ The angle is $$\theta=26.6^\circ$$
west of north.
$\textbf{(b)}$ The speed is
$$v_2=1.07\text{ m/s}.$$
Work Step by Step
$\textbf{(a)}$ The angle $\theta$ is the same as the angle at the 'upper' vertex of the triangle in the figure. Further, since the angle at the point $1$ is $45^\circ$, Then the altitude of the triangle splits the base of the length of $1.5$ km into two pieces, of $1.0$ km and $0.5$ km. Now we have
$$\tan\theta=\frac{0.5\text{ km}}{1.0\text{ km}}=0.5$$
so we get
$$\theta=\arctan 0.5 =26.6^\circ.$$
$\textbf{(b)}$ From the Pythagorean theorem we get that the distance that canoeist $2$ must cover is
$d_2=\sqrt{(1.0\text{ km})^2+(0.5\text{ km})^2} = \sqrt{1.25}\text{ km}=1.12\text{ km}.$
Similarly, the distance that the canoeist $1$ has to cover is
$d_1=\sqrt{(1.0\text{ km})^2+(1.0\text{ km})^2}=1.41\text{ km}.$
Equating the required travel times, which is simply distance over speed, we get:
$$\frac{d_1}{v_1}=\frac{d_2}{v_2}\Rightarrow v_2=\frac{d_2}{d_1}v_1=\frac{1.12\text{ km}}{1.41\text{ km}}\cdot1.35\text{ m/s}=1.07\text{ m/s}.$$
