Almost complex structures on spheres
Solution 1:
The point is that $V = \mathbb{R}^3 = \operatorname{Im}\mathbb{H}$ and $V = \mathbb{R}^7 = \operatorname{Im}\mathbb{O}$ inherit a cross-product $V \times V \to V$ from quaternion and octonion multiplication. It is given by $u \times v = \operatorname{Im}(uv) = \frac{1}{2}(uv - vu)$.
Given an oriented hypersurface $\Sigma \subset V$ with corresponding Gauß map $\nu \colon \Sigma \to S^n$ ($n \in 2,6$) sending a point $x \in \Sigma$ to the outer unit normal $\nu(x) \perp T_x\Sigma$ one obtains an almost complex structure by setting $$J_x(u) = \nu(x) \times u\qquad\text{for }u \in T_x\Sigma.$$
One economic way to see that this is an almost complex structure: The cross-product is bilinear and antisymmetric. It is related to the standard scalar product via $$ \langle u \times v, w\rangle = \langle u, v \times w\rangle $$ (which shows that $u\times v$ is orthogonal to both $u$ and $v$) and the Graßmann identity $u \times (v \times w) = \langle u,w\rangle v - \langle u,v\rangle w$ (only valid in dimension $3$) has the variant $$ (u \times v) \times w + u \times (v\times w) = 2\langle u,w\rangle v-\langle v,w\rangle u - \langle v,u\rangle w $$ valid in $3$ and $7$ dimensions (which shows that $u \times (u \times v) = -v$ for $u \perp v$ and $|u| = 1$).
Therefore $J_x(u) = \nu(x) \times u$ indeed defines an almost complex structure $J_x \colon T_x\Sigma \to T_x\Sigma$.
Specializing this to $\Sigma = S^2$ embedded as unit sphere in $\operatorname{Im}\mathbb{H}$ you can “see” (or calculate) that $J_x$ acts in exactly the same way as multiplication by $i$ on $\mathbb{CP}^1$.
Solution 2:
Yes and yes.
We get an almost complex structure $J$ on the two-sphere $S = \{ ai + bj + ck \mid a^2 + b^2 + c^2 = 1\}$ embedded in the space of imaginary quaternions in exactly the same way as for the octonions.
Since $S$ is of real dimension two, every almost complex structure on $S$ is integrable. Then $X = (S,J)$ is a compact complex curve of genus 0, and any such curve is isomorphic to $\mathbb{CP}^1$.