The tangent bundle of $S^m$ is known to be trivial only for $m=1,3,7$. Can this result be used to deduce that the tangent bundle of projective space $\mathbb{R}P^m$ is also trivial? (Clearly, for $n=1$ this works since $\mathbb{R}P^1\cong S^1$.) More generally:
For which positive integers $m$ is the tangent bundle $T\mathbb{R}P^m$ a trivial bundle?
One can ask the same question about the tangent bundles $T\mathbb{C}P^m$ of complex projective spaces.
A manifold is called parallelisable if it has trivial tangent bundle. So your question is for which positive integers $m$ is $\mathbb{RP}^m$ parallelisable? What about $\mathbb{CP}^m$?
If $T\mathbb{RP}^m$ is trivial, then the total Stiefel-Whitney class $w(T\mathbb{RP}^m)$ is $1$. By Corollary $4.6$ of Milnor & Stasheff's Characteristic Classes, $w(T\mathbb{RP}^m) = 1$ if and only if $m + 1$ is a power of $2$. So the only real projective spaces which can possibly be parallelisable are $\mathbb{RP}^1$, $\mathbb{RP}^3, \mathbb{RP}^7, \mathbb{RP}^{15}, \mathbb{RP}^{31}, \dots$ We still need to determine which of these are actually parallelisable (the condition on the total Stiefel-Whitney class is a necessary condition, but not sufficient as the case of spheres demonstrates).
Theorem $4.7$ of the same book states that if $\mathbb{R}^n$ admits a bilinear map $p : \mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}^n$ without zero divisors, then $\mathbb{RP}^{n-1}$ is parallelisable. By identifying $\mathbb{R}^2$ with $\mathbb{C}$, we obtain such a bilinear map for $n = 2$ given by the usual multiplication of complex numbers, i.e. $p((a, b),(c, d)) = (ac - bd, ac + bd)$. Similarly, by identifying $\mathbb{R}^4$ with the quaternions $\mathbb{H}$ and $\mathbb{R}^8$ with the octionions $\mathbb{O}$, we obtain such a map for $n = 4$ and $n = 8$. Therefore, $\mathbb{RP}^1$, $\mathbb{RP}^3$ and $\mathbb{RP}^7$ are parallelisable.
What about $\mathbb{RP}^{15}$, $\mathbb{RP}^{31}, \dots$? None of the remaining real projective spaces are parallelisable - this is harder to show. It follows from Kervaire's $1958$ paper Non-parallelizability of the $n$-sphere for $n > 7$.
So, the positive integers $m$ for which $\mathbb{RP}^m$ is parallelisable are $m = 1, 3,$ and $7$.
The parallelisability of $\mathbb{CP}^m$ is much easier to determine: $\mathbb{CP}^m$ is not parallelisable for any positive integer $m$.
As $\mathbb{CP}^m$ is a complex manifold, its tangent bundle is a complex vector bundle and hence has a total Chern class. The total Chern class of $\mathbb{CP}^m$ is $c(T\mathbb{CP}^m) = (1 + a)^{m+1} \in H^*(\mathbb{CP}^m, \mathbb{Z})$. If $T\mathbb{CP}^m$ were trivial, then it would have total Chern class $1$ but this is never the case. More explicitly,
$$c_1(T\mathbb{CP}^m) = (m+1)a \in H^2(\mathbb{CP}^m, \mathbb{Z}) \cong \mathbb{Z}.$$
As $a \neq 0$ and $m \neq -1$, $c_1(T\mathbb{CP}^m) = (m + 1)a \neq 0$.
More generally, any oriented closed manifold which is parallelisable must have Euler characteristic zero. In fact, such an oriented closed manifold which admits a nowhere zero vector field has Euler characteristic zero, see Property $9.7$ of the aforementioned book. The claim then follows from the fact that $\chi(\mathbb{CP}^m) = m + 1 \neq 0$.
When $m$ is even, the Euler characteristic of $\mathbb{RP}^m$ is $1$, so the top Stiefel-Whitney class of its tangent bundle doesn't vanish, and in particular its tangent bundle is not even stably trivial. In general, it's a nice exercise to calculate the Stiefel-Whitney classes of $\mathbb{RP}^m$, and the result you get is that they all vanish iff $m$ is one less than a power of $2$; in all other cases, the tangent bundle again fails to be stably trivial.
But we can do better. If $\mathbb{RP}^m$ has trivial tangent bundle then so does its double cover $S^m$, and this is known to happen iff $m = 1, 3, 7$. It's clear that $\mathbb{RP}^1 \cong S^1$ and $\mathbb{RP}^3 \cong SO(3)$ have trivial tangent bundles; I'm less clear on $\mathbb{RP}^7$.
The argument is even simpler for $\mathbb{CP}^m$: $\mathbb{CP}^m$ is both always orientable and always has nonzero Euler characteristic (namely $m + 1$), so the Euler class of its (oriented) tangent bundle doesn't vanish. Equivalently, by the converse of the Poincare-Hopf theorem, it doesn't admit a nonvanishing vector field.
