I am wondering if there is a standard $N$-sphere that has non-trivial first Pontryagin class on its tangent bundle $TS^n$and frame bundle $FS^n$?
I know that only $S^4$ has non-trivial $H^4(S^n, R)$ cohomology class, while locally I write the first Pontryagin class on $TS^n$ as $Tr(R^2)$, where $R$ is curvature of $S^n$, is trivially zero. Does it mean all sphere have trivial first Pontryagin class?
On
All positive characteristic classes vanish for the tangent bundles of standard spheres, as they are stably trivial. That is, $TS^{n} \oplus \mathbb{R}$ is a trivial bundle. One might see it by looking at the standard embedding $S^{n} \hookrightarrow \mathbb{R}^{n+1}$, where we have
$\mathbb{R}^{n+1} \simeq T\mathbb{R}^{n+1} |_{S^{n}} \simeq N_{\mathbb{R}^{n+1} / S^{n+1}} \oplus TS^{n+1}$.
Now, the normal bundle is trivial as one can orient it consistently in either the "zero-pointing" or "infinity-pointing" directions in $\mathbb{R}^{n+1}$. To see that this implies your result, recall that the Pontryagin classes are the Chern classes of complexification and we have
$\mathbb{C}^{n+1} \simeq (TS^{n} \oplus \mathbb{R}) \otimes \mathbb{C} \simeq (TS^{n} \otimes \mathbb{C}) \oplus \mathbb{C}$.
The Whitney formula implies that
$1 = c(\mathbb{C}^{n+1}) = c((TS^{n} \otimes \mathbb{C}) \oplus \mathbb{C}) \simeq c(TS^{n} \otimes \mathbb{C}) * c(\mathbb{C}) = c(TS^{n} \otimes \mathbb{C})$,
so all the Pontryagin classes vanish.
Since $p_1(S^n) \in H^4(S^n; \Bbb Z)$, the only possible sphere $S^n$ with $p_1(S^n) \neq 0$ is $S^4$. Now $S^4$ can be considered as the quaternionic projective line $\Bbb H P^1$. There is a general formula for the total Pontryagin class of $\Bbb H P^m$: $$p(\Bbb H P^m) = \frac{(1+u)^{2m+2}}{1+4u},$$ where $u$ is a generator of $H^4(\Bbb H P^m; \Bbb Z)$ (see Exercise 20-A of Milnor and Stasheff's Characteristic Classes). It follows that $$p_1(\Bbb H P^m) = (2m-2)u.$$ Therefore $$p_1(S^4) = p_1(\Bbb H P^1) = 0,$$ and all (standard) spheres have trivial first Pontryagin class.