Answer
$\textbf{(a)}$ $h=\frac{v_v^2}{2g}=3.1\text{ m}.$
$\textbf{(b)}$ $l=6.2\text{ m}$ from the edge of the first cliff, i.e. $l'=3.4\text{ m}$ away from the second cliff.
Work Step by Step
$\textbf{(a)}$ The fact that the angle of landing is $-45^\circ$ (i.e. $45^\circ$ below the horizontal) means that the vertical component of the velocity and the horizontal component of the velocity are equal in magnitude. This means that the magnitude of the vertical component at the landing is $v_v=7.8\text{ m/s}$. Since the initial vertical component is equal to zero, we have for the height difference:
$$v_v^2=2gh\Rightarrow h=\frac{v_v^2}{2g}=\frac{7.8^2(\text{m/s})^2}{2\cdot9.81\text{ m/s}^2} = 3.1\text{ m}.$$
$\textbf{(b)}$ The climber lands on the other side of the crevasse. We will now calculate the horizontal distance he covers. The time of the descent is given by:
$$v_v=gt\Rightarrow t=\frac{v_v}{g}.$$
Now, since the horizontal component of the velocity is constant, the horizontal distance covered is
$$l=v_ht=v_h\frac{v_v}{g}=\frac{v_h^2}{g}=\frac{7.8^2(\text{m/s})^2}{9.81\text{ m/s}^2}=6.2\text{ m},$$
where we used the fact from part $\textbf{(a)}$ that $v_v=v_h=7.8\text{ m/s}.$
So, the climber lands
$$l'=l-2.8\text{ m}=3.4\text{ m}$$
away from the edge of the second cliff.
