No. There is probably a straightforward representation-theoretic argument, but I am too ignorant of the subject to give one, so here is a topological argument.

If $H \subset G$ are Lie groups, then $G/H$ has the natural structure of a smooth manifold without boundary. If $G$ is compact, so is $G/H$, as it is the continuous image of a compact space.

Now $\dim SU_3 = 8$ and $\dim SO_4 = 6$. Therefore, if there existed some group embedding $j: SO_4 \hookrightarrow SU_3$, the quotient $X = SU_3/j(SO_4)$ must be a compact 2-manifold without boundary.

Furthermore, the fibering $SO_4 \xrightarrow{j} SU_3 \to X$ induces the long exact sequence of homotopy groups $$\pi_2 SU_3 \to \pi_2 X \to \pi_1 SO_4 \to \pi_1 SU_3 \to \pi_1 X \to \pi_0 SO_4.$$ Filling in the values we know for these groups, we find that there is a long exact sequence of groups $$0 \to \pi_2 X \to \Bbb Z/2 \to 0 \to \pi_1 X \to 0.$$ Therefore $\pi_2 X \cong \Bbb Z/2$ and $\pi_1 X \cong 0$. This is a contradiction; the only simply connected compact surface without boundary is the 2-sphere, which has $\pi_2 X = \Bbb Z$.

No. There is probably a straightforward representation-theoretic argument, but I am too ignorant of the subject to give one, so here is a topological argument.

If $H \subset G$ are Lie groups with $H$ closed in $G$, then $G/H$ has the natural structure of a smooth manifold without boundary. If $G$ is compact, so is $G/H$, as it is the continuous image of a compact space; if $H$ is compact the closed-ness condition is automatic.

Now $\dim SU_3 = 8$ and $\dim SO_4 = 6$. Therefore, if there existed some group embedding $j: SO_4 \hookrightarrow SU_3$, the quotient $X = SU_3/j(SO_4)$ must be a compact 2-manifold without boundary.

Furthermore, the fibering $SO_4 \xrightarrow{j} SU_3 \to X$ induces the long exact sequence of homotopy groups $$\pi_2 SU_3 \to \pi_2 X \to \pi_1 SO_4 \to \pi_1 SU_3 \to \pi_1 X \to \pi_0 SO_4.$$ Filling in the values we know for these groups, we find that there is a long exact sequence of groups $$0 \to \pi_2 X \to \Bbb Z/2 \to 0 \to \pi_1 X \to 0.$$ Therefore $\pi_2 X \cong \Bbb Z/2$ and $\pi_1 X \cong 0$. This is a contradiction; the only simply connected compact surface without boundary is the 2-sphere, which has $\pi_2 X = \Bbb Z$.

No. There is probably a straightforward representation-theoretic argument, but I am too ignorant of the subject to give one, so here is a topological argument.

If $H \subset G$ are Lie groups, then $G/H$ has the natural structure of a smooth manifold without boundary. If $G$ is compact, so is $G/H$, as it is the continuous image of a compact space.

Now $\dim SU_3 = 8$ and $\dim SO_4 = 6$. Therefore, if there existed some group embedding $j: SO_4 \hookrightarrow SU_3$, the quotient $X = SU_3/j(SO_4)$ must be a compact 2-manifold without boundary.

Furthermore, the fibering $SO_4 \xrightarrow{j} SU_3 \to X$ induces the long exact sequence of homotopy groups $$\pi_2 SU_3 \to \pi_2 X \to \pi_1 SO_4 \to \pi_1 SU(3) \to \pi_1 X \to \pi_0 SO_4.$$ Filling in the values we know for these groups, we find that there is a long exact sequence of groups $$0 \to \pi_2 X \to \Bbb Z/2 \to 0 \to \pi_1 X \to 0.$$ Therefore $\pi_2 X \cong \Bbb Z/2$ and $\pi_1 X \cong 0$. This is a contradiction; the only simply connected compact surface without boundary is the 2-sphere, which has $\pi_2 X = \Bbb Z$.

No. There is probably a straightforward representation-theoretic argument, but I am too ignorant of the subject to give one, so here is a topological argument.

If $H \subset G$ are Lie groups, then $G/H$ has the natural structure of a smooth manifold without boundary. If $G$ is compact, so is $G/H$, as it is the continuous image of a compact space.

Now $\dim SU_3 = 8$ and $\dim SO_4 = 6$. Therefore, if there existed some group embedding $j: SO_4 \hookrightarrow SU_3$, the quotient $X = SU_3/j(SO_4)$ must be a compact 2-manifold without boundary.

Furthermore, the fibering $SO_4 \xrightarrow{j} SU_3 \to X$ induces the long exact sequence of homotopy groups $$\pi_2 SU_3 \to \pi_2 X \to \pi_1 SO_4 \to \pi_1 SU_3 \to \pi_1 X \to \pi_0 SO_4.$$ Filling in the values we know for these groups, we find that there is a long exact sequence of groups $$0 \to \pi_2 X \to \Bbb Z/2 \to 0 \to \pi_1 X \to 0.$$ Therefore $\pi_2 X \cong \Bbb Z/2$ and $\pi_1 X \cong 0$. This is a contradiction; the only simply connected compact surface without boundary is the 2-sphere, which has $\pi_2 X = \Bbb Z$.