Lie group representation, exponential, and $\theta$-periodicty 
SU(2) 

I know we can view the group element in the SU(2) Lie group as
$$ g = \exp\left(\theta\sum_{k=1}^{3} i t_k \frac{\sigma_k}{2}\right)  $$
where $(t_1,t_2,t_3)$ forms a unit vector [effectively pointing in some direction on a unit 2-sphere $S^2$], and $\sigma_k$ are Pauli matrices:
\begin{align}
  \sigma_1 = \sigma_x &=
    \begin{pmatrix}
      0&1\\
      1&0
    \end{pmatrix} \\
  \sigma_2 = \sigma_y &=
    \begin{pmatrix}
      0&-i\\
      i&0
    \end{pmatrix} \\
  \sigma_3 = \sigma_z &=
    \begin{pmatrix}
      1&0\\
      0&-1
    \end{pmatrix} \,.
\end{align}
Notice that any group element on $SU(2)$ can be parametrized by some $\theta$ and $(t_1,t_2,t_3)$. Also $\theta$ has a periodicity $[0,4 \pi)$, instead of $2 \pi$.
Notice that in this case we also have
$$ g = \exp\left(\theta\sum_{k=1}^{3} i t_k \frac{\sigma_k}{2}\right)  
=\cos(\frac{\theta}{2})+i \sum_{k=1}^{3}  t_k \sigma_k\sin(\frac{\theta}{2})$$

question 1: SU(3) 

(1) Is this true that all $SU(3)$ group elements can be written as:
$$ g = \exp\left(\theta\sum_{k=1}^{8} i t_k \frac{\lambda_k}{2}\right)=\cos(\frac{\theta}{2})+i \sum_{k=1}^{8}  t_k \lambda_k\sin(\frac{\theta}{2})  $$
where $\lambda_k$ are Gell-Mann_matrices? And does the second equality still hold? Here Tr$(\lambda_k^2)=2.$
Also $\theta$ has a periodicity $[0,4 \pi)$?

question 2: SU(n), for $n=4, ...$ 

(2) Is this true that all $SU(4)$ group elements can be written as:
$$ g = \exp\left(\theta\sum_{k=1}^{4^2-1} i t_k \frac{\lambda_k}{2}\right)=\cos(\frac{\theta}{2})+i \sum_{k=1}^{4^2-1}  t_k \lambda_k\sin(\frac{\theta}{2})  $$
where $\lambda_k$ are generalized rank-4 Gell-Mann_matrices in eq.(3)? And does the second equality still hold? Here Tr$(\lambda_k^2)=2.$
Also $\theta$ has a periodicity $[0,4 \pi)$?
(3) How to determine the Right Hand side equation and $\theta$  periodicity?
 A: There are two key facts that contribute the SU(2) formula for Pauli matrices:

*

*Distinct Pauli matrices anticommute

*Times them by $i$ to get square roots of $-I$
The importance can be seen by squaring a "vector" with them as basis:
$$ (it_1\sigma_1+it_2\sigma_2+it_3\sigma_3)^2=-(t_1^2+t_2^2+t_3^2) $$
(working this algebra out is instructive). This means $t:=it_1\sigma_1+it_2\sigma_2+it_3\sigma_3$ is a square root of $-I$ when $(t_1,t_2,t_3)$ is a unit vector, and thus by the usual proof of de Moivre's formula,
$$ \exp(\theta t)=\cos(\theta)I+\sin(\theta)t. $$
However the Gell-Mann matrices are not as nice. First notice
$$ (i\lambda_1)^2=(i\lambda_2)^2=\mathrm{diag}(-1,-1,0) \\ (i\lambda_4)^2=(i\lambda_5)^2=\mathrm{diag}(-1,0,-1) \\ (i\lambda_6)^2=(i\lambda_7)^2=\mathrm{diag}(0,-1,-1) $$
so these are not quite square roots of $-I$, although that could potentially be fixed in a tentative exponential formula since their sum is a multiple of $-I$. But then the symmetry gets broken:
$$ (i\lambda_3)^2 = \mathrm{diag}(-1,-1,0) \\ (i\lambda_8)^2=\mathrm{diag}(-\frac{1}{3},-\frac{1}{3},-\frac{4}{3}). $$
Thus, in particular,
$$\exp(\theta i\lambda_8)\ne \cos(\theta)+i \sin(\theta)\lambda_8.$$
Moreover, one can verify some of the pairs of Gell-Mann matrices do not anticommute.
A: At least for SU(3), the exponential of arbitrary Lie-algebra elements in the fundamental representation is a bit more complicated, and, as a consequence of the Cayley-Hamilton theory, it also has a term bilinear in the Lie algebra elements, unlike the simple SU(2) expression you may be hankering for. See Curtright & Zachos 2015, and arXiv.
Specifically, the generic  SU(3) group element generated by a traceless 3×3 Hermitian matrix  H, normalized as  trH2 = 2, can be expressed as a second order  matrix polynomial in  H,
$$\begin{align}
  \exp(i\theta H) ={}
          &\left[-\frac{1}{3} I\sin\left(\varphi + \frac{2\pi}{3}\right) \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H\sin(\varphi) - \frac{1}{4}~H^2\right]
           \frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta\sin(\varphi)\right)}
                {\cos\left(\varphi + \frac{2\pi}{3}\right) \cos\left(\varphi - \frac{2\pi}{3}\right)} \\[4pt]
    {}+{} &\left[-\frac{1}{3}~I\sin(\varphi) \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H\sin\left(\varphi + \frac{2\pi}{3}\right) - \frac{1}{4}~H^{2}\right]
           \frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta \sin\left(\varphi + \frac{2\pi}{3}\right)\right)}
                {\cos(\varphi) \cos\left(\varphi - \frac{2\pi}{3}\right)} \\[4pt]
    {}+{} &\left[-\frac{1}{3}~I\sin(\varphi) \sin\left(\varphi + \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{4}~H^2\right]
           \frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta \sin\left(\varphi - \frac{2\pi}{3}\right)\right)}
                {\cos(\varphi)\cos\left(\varphi + \frac{2\pi}{3}\right)}
\end{align} $$
where
 $$\varphi \equiv \frac{1}{3}\left[\arccos\left(\frac{3\sqrt{3}}{2}\det H\right) - \frac{\pi}{2}\right]. $$
