Computation of $\pi_4$ of simple Lie groups Below we assume any simple Lie group $G$ to be simply connected.
$\pi_3(G)=\mathbb{Z}$ for any simple Lie group $G$ and there is a uniform proof for that.
Now the textbooks say $\pi_4(G)$ is trivial except for $G=Sp(n)$, for which it is $\mathbb{Z}/2\mathbb{Z}$. 
My question is the following: is there a uniform way to derive this fact, in such a way to show which property of the symplectic group is so special?
 A: There is actually a collection of more general statements that detail the structure of the $\pi_4$.
Firstly it was Browder [The Cohomology of Covering Spaces of H-Spaces] who proved that a simply connected finite H-space is 2-connected. Clark [On $\pi_3$ of Finite Dimensional H-Spaces] showed that the first non-vanishing homotopy group of a finite loop space appeared in degrees 1 or 3 (the result that $\pi_2=0$ for a simply connected finite H-space also appears in Browders [Torsion in H-Spaces], see also Lin's [The First Homotopy Group of a Finite H-Space]). Kane and Hubbuck [On $\pi_3$ of a Finite H-Space] prove that $\pi_3$ is torsion free, building on previous work of Thomas and Lin which required stricter assumptions (the Lie group $E_8$ does not fall under Thomas's requirements, for instance).
Finally Hubbuck [Homotopy Groups of Finite H-Spaces] studied $\pi_4$, showing that it was a direct sum of finite groups of order 2. The proof uses JHC Whitehead's Certain Exact Sequence (see Baues [Homotopy Type and Homology] for a good discussion), identifying the groups that appear within using the work of others. The end result is an exact sequence
$H_5(X;\mathbb{Z})\xrightarrow{\rho_2} H_5(X;\mathbb{Z}_2)\xrightarrow{Sq^2_*}H_3(X;\mathbb{Z}_2)\rightarrow \pi_4X\rightarrow0$
And we know that $H_3(X;\mathbb{Z}_2)=\oplus\,\mathbb{Z}_2$ from the Hurewicz theorem and the work of Kane and Hubbuck so the result on $\pi_4X$ follows.
Harper claims that if $X$ is simply connected and $H_*(\Omega X;\mathbb{Z})$ is torsion free then the result follows from the previous work of Bott and Samelson [Applications of the Theory of Morse to Symmetric Spaces].
A: As a partial answer, here's at least a uniform statement, which can be found as Theorem 3.10 in Mimura's survey "Homotopy theory of Lie groups" in the Handbook of Algebraic Topology: 

Let $G$ be a compact, connected, simply connected, simple Lie group, $T$ a maximal torus of $G$, $\mathbb{R}^r$ its universal covering and $\Gamma$ the inverse image of the identity in $\mathbb{R}^r$. Let $a$ be the dominant root with respect to some lexicographic order of the roots of $G$. Then $\pi_4(G)$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ if the hyperplane $a=1$ contains a point of $\Gamma$ and $0$ otherwise. 

(In the survey article, the statement has a $\mathbb{Z}$ where I wrote $\mathbb{Z}/2\mathbb{Z}$ but that would appear to be a typo.)
Mimura refers to the paper


*

*R. Bott and H. Samelson. Applications of the theory of Morse to symmetric spaces. Amer. J. Math. 80 (1958), 964-1029. 


I can't check right now, but the uniform statement suggests a uniform proof, relating the fourth homotopy group to information on the root system. It would seem that the difference that singles out the symplectic groups is that the long roots are divisible (by 2 incidentally) in the character group.
A: Since you are asking a question about $\pi_4$, lets use a little homotopy theory to think about this, and for starters, we can just ponder compact Lie groups.  
Consider first $G=Sp(1)=SU(2)=S^3$.  It is well known that $\pi_4(S^3) = \mathbb Z/2$.  
Now lets compare and contrast $SU(3)$ with $Sp(2)$.  $SU(3)/SU(2) = S^5$, while $Sp(2)/Sp(1) = S^7$, and these lead to fibration sequences
$ S^3 \rightarrow SU(3) \rightarrow S^5$
and 
$ S^3 \rightarrow Sp(2) \rightarrow S^7$.   The associated long exact sequences of homotopy groups make it obvious that $\pi_4(Sp(2)) = \mathbb Z/2$; more surprising from this perspective is that $\pi_4(SU(3)) = 0$.
In general, well chosen (= highly connected) homogeneous spaces $G/H$ let one get at $\pi_*(G)$ from knowledge of $\pi_*(H)$ for $*$ not too large.  For example, $G_2/SU(3) = S^6$, so $\pi_4(G_2) = 0$.
Finally, here is a general method:  Let $G\langle 3\rangle$ be the fiber of the fundamental class $G \rightarrow K(\mathbb Z,3)$.  Then $\pi_4(G) = \pi_4(G\langle 3\rangle) = H_4(G\langle 3\rangle;\mathbb Z)$.  Then one can try to get at this homology group using the Serre spectral sequence associated to the fibration $K(\mathbb Z,2) \rightarrow G\langle 3 \rangle \rightarrow G$.   
In various papers, M. Mimura has done much analysis of this sort, and in particular computes the first couple dozen homotopy groups of all of the exceptional Lie groups.
