= Solution
The <SU(2) group> consists of complex $2\times2$ matrices $U$ with $U^\dagger U=I$ and $\det U=1$. The <SO(3) group> consists of real $3\times3$ matrices $R$ with $R^TR=I$ and $\det R=1$. Differentiate these equations along a path through the identity. This gives
$$
\mathfrak{su}(2)=\{X:X^\dagger=-X,\ \operatorname{tr}X=0\},\qquad
\mathfrak{so}(3)=\{L:L^T=-L\}.
$$
Conversely the <matrix exponential> of each displayed infinitesimal matrix satisfies the corresponding group equations, so these are exactly the <Lie algebras>, with bracket the matrix <commutator>.
Using the <Pauli matrices>, put $t_a=-i\sigma_a/2$. They form a real basis of the <SU(2) Lie algebra>. The <Pauli matrix commutator identity> gives $[t_a,t_b]=\epsilon_{abc}t_c$. Define $J_a$ on $\mathbb R^3$ by $J_av=e_a\times v$. These are a basis of the <SO(3) Lie algebra>; the vector triple-product identity gives $[J_a,J_b]=\epsilon_{abc}J_c$. Thus
$$
\boxed{\sum_a u_at_a\longmapsto\sum_a u_aJ_a}
$$
is a real linear bijection preserving the bracket. This is the <SU(2)-SO(3) Lie algebra isomorphism>. It is a Lie-algebra isomorphism, not a group isomorphism: the <Adjoint double cover from SU(2) to SO(3)> has kernel $\{I,-I\}$.
The <SU(3) group> is defined similarly by $U^\dagger U=I$ and $\det U=1$ on complex $3\times3$ matrices. One <SU(2)> subgroup is $\{\operatorname{diag}(V,1):V\in SU(2)\}$. The real orthogonal matrices with determinant one form an <SO(3) group> subgroup, since real orthogonality is also complex unitarity.
Under the defining <group action>, the <group orbit> of $e_1$ is exactly the unit sphere in $\mathbb C^3$:
$$
\boxed{SU(3)e_1=\{z:z^\dagger z=1\}=S^5.}
$$
Unitarity proves containment. Conversely, extend a unit vector $z$ to an orthonormal basis and use it as the first column of a unitary matrix. Multiplying the last column by the inverse of its determinant makes that determinant one without changing the first column. The <isotropy group> of $e_1$ must also preserve its orthogonal complement, and therefore is
$$
\boxed{\{\operatorname{diag}(1,V):V\in SU(2)\}\cong SU(2).}
$$
This proves the <unit sphere orbit of the defining special unitary action>, including $S^5\cong SU(3)/SU(2)$.
For the complex quadric, write $z=x+iy$. Its defining relation separates into
$$
|x|^2-|y|^2=1,\qquad x\cdot y=0.
$$
The Hermitian norm is then $z^\dagger z=1+2|y|^2$, which is not constant on this set. For example, $e_1$ and $\sqrt2e_1+ie_2$ both satisfy the complex bilinear relation but have Hermitian norms one and three. Since the <SU(3) group> preserves that norm, \b[the quadric cannot be a single SU(3) <group orbit>]. In fact it is not even invariant under the whole group: $\operatorname{diag}(e^{it},e^{-it},1)e_1$ fails the bilinear relation when $e^{2it}\ne1$.
The real <SO(3) group> does preserve the quadric, acting simultaneously on $x$ and $y$. Its <group orbits> are classified completely by $r=|y|\geq0$. For $r=0$, $y=0$ and $x$ is a real unit vector, giving the <group orbit> $S^2$ and <stabilizer subgroup> $SO(2)$. For $r>0$, the vectors $x/\sqrt{1+r^2}$ and $y/r$ are orthonormal. Adjoining their cross product gives an oriented orthonormal frame. There is a unique rotation taking the standard frame to this one; hence the action is transitive at fixed $r$ and the <stabilizer subgroup> is trivial. Thus the <real rotation orbits on a complex unit quadric> are
$$
\boxed{M_0\cong S^2,\qquad M_r\cong SO(3)\ (r>0),\qquad M/SO(3)\cong[0,\infty).}
$$
In particular the real subgroup does not act transitively on the whole quadric. The parametrization $x=\sqrt{1+|y|^2}\,n$ with $n\in S^2$ and $y\perp n$ also identifies the quadric, as a real manifold, with the <tangent bundle> of $S^2$.
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