= Solution
A <maximal torus> in a compact connected <Lie group> is a compact connected abelian Lie subgroup maximal among such subgroups. By the connected abelian classification it is isomorphic to a product of circles. Such subgroups exist: choose one of largest dimension among compact connected abelian subgroups; a proper inclusion of connected Lie subgroups would increase dimension. Write $\mathfrak t$ for its <Lie algebra>.
Here are the main steps of the conjugacy assertion, with the critical arguments supplied. Average an inner product on $\mathfrak g$ over the adjoint action to obtain an $\operatorname{Ad}(G)$-invariant positive definite inner product. It defines a <bi-invariant Riemannian metric>. Compactness makes this metric complete, and the <Hopf-Rinow theorem> joins the identity to any $g$ by a geodesic. For a bi-invariant metric the Levi-Civita connection on left-invariant fields is $\nabla_XY=[X,Y]/2$, so the geodesic starting with $X$ is $\exp(tX)$. Consequently every $g$ has the form $\exp X$.
The Lie-algebra centralizer of $\mathfrak t$ is exactly $\mathfrak t$. If $X$ commutes with $\mathfrak t$, its <one-parameter subgroup> commutes with $T$. The closure of the subgroup generated by $T$ and $\exp(\mathbb RX)$ is compact, connected and abelian, so it is a torus; maximality forces it to equal $T$, giving $X\in\mathfrak t$. The torus action on $\mathfrak g_\mathbb C$ decomposes into finitely many <character> <weight spaces>. The zero-<weight space> is $\mathfrak t_\mathbb C$. Choose $H\in\mathfrak t$ outside the finitely many kernels of the nonzero weight differentials. Then the centralizer of $H$ is exactly $\mathfrak t$.
For an arbitrary $X\in\mathfrak g$, the <continuous> function $g\mapsto\langle\operatorname{Ad}_gX,H\rangle$ attains a maximum. Put $Y=\operatorname{Ad}_{g_0}X$ at a maximizer. Differentiating along $\exp(sZ)g_0$ gives
$$
0=\langle[Z,Y],H\rangle=\langle Z,[Y,H]\rangle
\quad\text{for every }Z\in\mathfrak g.
$$
Hence $[Y,H]=0$, so $Y\in\mathfrak t$. Exponentiating gives $g_0(\exp X)g_0^{-1}=\exp Y\in T$. This <adjoint-orbit criterion for maximal-torus conjugacy> proves
$$
\boxed{\text{every element of }G\text{ belongs to a subgroup conjugate to }T.}
$$
Restriction of <representations> is a <ring homomorphism> $\operatorname{Res}:R(G)\to R(T)$. If a virtual <representation> restricts to zero, its <character> vanishes on $T$. Since <characters> are <class functions> and every element is conjugate into $T$, it vanishes on all of $G$. The injectivity of the <character> map proved in Question 3 then gives
$$
\boxed{\operatorname{Res}:R(G)\longrightarrow R(T)\text{ is injective}.}
$$
The <Weyl group> is $W=N_G(T)/T$. For $n\in N_G(T)$, the operator $\rho(n)$ intertwines the restriction of $\rho$ with its twist by $t\mapsto ntn^{-1}$. Therefore every restricted <representation>, and every virtual difference of them, is fixed by $W$:
$$
\boxed{\operatorname{Res}(R(G))\subseteq R(T)^W.}
$$
For the <special orthogonal group> $G=SO(2m+1)$, take
$$
T=\{\operatorname{diag}(R_{\theta_1},\ldots,R_{\theta_m},1)\},
$$
where $R_\theta$ is a real planar rotation. It is a <maximal torus>: its Lie-algebra centralizer consists of the independent infinitesimal rotations in those planes. Its <character> lattice is $\mathbb Z^m$, so, writing $z_i=e^{i\theta_i}$,
$$
R(T)=\mathbb Z[z_1^{\pm1},\ldots,z_m^{\pm1}].
$$
Indeed commuting unitary torus <matrices> diagonalize simultaneously into one-dimensional <characters>, and the <characters> of the circle are exactly the integral powers.
The normalizer permutes the real rotation planes and can reverse each angle independently. Permutations of whole two-dimensional blocks have determinant one. A reflection in one rotation plane reverses its angle and has determinant minus one; combining it with reflection of the remaining fixed line gives an element of $SO(2m+1)$. Thus all independent inversions occur. Conversely a normalizer must permute the distinct nontrivial torus <character> pairs $\{\pm e_i\}$ and preserve the fixed line, so there are no additional actions. The kernel of this action is exactly $T$, giving
$$
\boxed{W\cong(\mathbb Z/2)^m\rtimes S_m,\qquad
z_i\mapsto z_{\sigma(i)}^{\pm1}.}
$$
It follows that the restriction identifies $R(SO(2m+1))$ with a subring of the signed-permutation-invariant <Laurent polynomials>. In this case we can prove equality. Invariance under individual inversions first gives
$$
\mathbb Z[z_1^{\pm1},\ldots,z_m^{\pm1}]^{(\mathbb Z/2)^m}
=\mathbb Z[x_1,\ldots,x_m],\qquad x_i=z_i+z_i^{-1}.
$$
For one variable, every invariant is an integral combination of $1$ and $z^k+z^{-k}$, and these satisfy the recurrence $s_{k+1}=x s_k-s_{k-1}$. Applying this one coordinate at a time proves the assertion over the other Laurent coefficient rings. Permutation invariance and the <Fundamental theorem of symmetric polynomials> then give
$$
R(T)^W=\mathbb Z[e_1(x),\ldots,e_m(x)],
$$
where $e_k$ are the <elementary symmetric polynomials>.
Let $V$ be the complexification of the defining $(2m+1)$-dimensional real <representation>. Its torus weights are $1,z_i,z_i^{-1}$. The <characters> of its exterior powers obey
$$
\sum_{k=0}^{2m+1}\chi_{\Lambda^kV}t^k
=(1+t)\prod_{i=1}^m(1+z_it)(1+z_i^{-1}t)
=(1+t)\prod_{i=1}^m(1+x_it+t^2).
$$
For $k\leq m$, choosing $j$ linear terms and $r$ quadratic terms shows that the coefficient $c_k=\chi_{\Lambda^kV}$ is
$$
c_k=
\sum_{j+2r=k}\binom{m-j}{r}e_j+
\sum_{j+2r=k-1}\binom{m-j}{r}e_j,\qquad e_0=1.
$$
Its leading term is $e_k$ with coefficient one; all other terms use $e_j$ with $j<k$. By induction $e_k$ lies in the restriction image, because $c_k$ is the <character> of a genuine <representation> and all smaller $e_j$ already lie in the image. Hence every invariant polynomial lies there, proving the <representation ring of an odd-dimensional special orthogonal group> description
$$
\boxed{R(SO(2m+1))\cong R(T)^W
=\mathbb Z[e_1,\ldots,e_m]
=\mathbb Z[c_1,\ldots,c_m].}
$$
The triangular change of generators is integral, so $[\Lambda^1V],\ldots,[\Lambda^mV]$ are algebraically independent polynomial generators. For $m=1$, $c_1=1+z+z^{-1}$ and this reduces to $R(SO(3))=\mathbb Z[V_3]$, with the three-dimensional defining <representation> as generator. These are genuine <representations> of the orthogonal <group>, not the half-integral spin weights of its simply connected double cover.
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