= Solution
A <group homomorphism> sends identity to identity. For $\theta(g),\theta(h)$ in its image,
$$
\theta(g)\theta(h)^{-1}=\theta(gh^{-1}),
$$
which is still in the image; hence the <image of a group homomorphism> is a <subgroup> of $H$. Let $N=\ker\theta$. It is a <subgroup> by the same identity, and for $k\in N$ and $g\in G$,
$$
\theta(gkg^{-1})=\theta(g)\theta(k)\theta(g)^{-1}=e.
$$
Thus $gNg^{-1}\subseteq N$, and replacing $g$ by $g^{-1}$ gives equality, proving that $N$ is a <normal subgroup>.
Define a map from the <quotient group> by
$$
\Phi:G/N\longrightarrow\theta(G),\qquad \Phi(gN)=\theta(g).
$$
If $gN=hN$, then $h^{-1}g\in N$ and $\theta(g)=\theta(h)$, so the map is well defined. Conversely equality of the images implies $h^{-1}g\in N$, giving injectivity. It is onto by the definition of image, and $\Phi((gN)(hN))=\theta(gh)=\theta(g)\theta(h)$. This proves the <first isomorphism theorem for groups> in this case:
$$
\boxed{G/\ker\theta\cong\theta(G)}.
$$
An <orthogonal matrix> $Q$ has $Q^{\mathsf T}Q=I$, so $(\det Q)^2=1$. The <determinant> is therefore a <group homomorphism> $O(3)\to\{1,-1\}$, with image both signs because $\operatorname{diag}(-1,1,1)$ has determinant $-1$. Its kernel is the <special orthogonal group> $SO(3)$. Hence \b[$SO(3)$ is normal in $O(3)$], and
$$
\boxed{O(3)/SO(3)\cong\{1,-1\}\cong C_2}.
$$
For the last request, the <determinant-twist projection in odd dimension> gives an explicit example:
$$
\boxed{\pi:O(3)\longrightarrow SO(3),\qquad \pi(Q)=(\det Q)Q}.
$$
Indeed $\det\pi(Q)=(\det Q)^4=1$, and scalar factors commute, so $\pi(QR)=\pi(Q)\pi(R)$. If $\pi(Q)=I$, then $Q=(\det Q)I$, making $Q=I$ or $Q=-I$. Both lie in the kernel, which is therefore exactly $\{I,-I\}$, of order two. The map also fixes every member of $SO(3)$, so it is onto.
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