Periodicity of the discrete-time Fourier Transform
The DTFT is always $2\pi$-periodic. However, it can also have a smaller period, namely a fraction of $2\pi$. Take any sequence $x[n]$ for which the DTFT exists and insert $L-1$ zeros between the samples. The DTFT of the new sequency $\hat{x}[n]$ can then be written as

$$\hat{X}(e^{j\omega})=\sum_{n=-\infty}^{\infty}\hat{x}[n]e^{-jn\omega}=\sum_{n=-\infty}^{\infty}\hat{x}[nL]e^{-jnL\omega}\tag{1}$$

$\hat{X}(e^{j\omega})$ as given by $(1)$ clearly has a period of $2\pi/L$.

Concerning your second question, the reason why a continuous-time Fourier transform $X(j\omega)$ cannot be written as $X(e^{j\omega})$ is that

$$X(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\neq \int_{-\infty}^{\infty}x(t)(e^{j\omega})^{-t}dt\tag{2}$$

simply because generally

$$e^{-j\omega t}\neq (e^{j\omega})^{-t}\tag{3}$$

unless $t$ is an integer (as is the case with the DTFT).

Note that if $(3)$ were not true,  i.e., if $e^{-j\omega t}= (e^{j\omega})^{-t}$ were true, we could easily show that $e^{-j\omega t}=1$ for all $t$. Just write $\omega=2\pi/T$ and you get

$$e^{-j\omega t}=e^{-j2\pi t/T}=\left(e^{-j2\pi}\right)^{t/T}=1^{t/T}=1$$

which is of course absurd.


Proof of Eq. $(3)$:

Let $z$ be a complex number with $|z|=1$ (the magnitude is irrelevant here):

$$z=e^{j\theta}\tag{4}$$

Let $a$ and $b$ be real numbers. Then

$$z^{ab}=e^{j\theta a b}\tag{5}$$

And with $z^a=u$

$$\left(z^a\right)^b=u^b=e^{j\arg\{u\}b}\tag{6}$$

With $\arg\{u\}=\text{pv}\{\theta a\}\in (-\pi,\pi]$, where $\text{pv}$ denotes the principal value, it is clear that generally

$$z^{ab}=e^{j\theta a b}\neq \left(z^a\right)^b=e^{j\,\text{pv}\{\theta a\}b}\tag{7}$$

There are two cases where $(5)$ and $(6)$ are equal:


if $\text{pv}\{\theta a\}=\theta a$, which is the case if $\theta a\in (-\pi,\pi]$.
if $b$ is integer, since


$$\text{pv}\{\theta a\}=\theta a+2\pi k\tag{8}$$

with some appropriately chosen integer $k$ (such that the result is in the interval $(-\pi,\pi]$), then

$$\left(z^a\right)^b=e^{j\,\text{pv}\{\theta a\}b}=e^{j(\theta a+2\pi k)b}=e^{j\theta ab}=z^{ab},\qquad b\in\mathbb{Z}\tag{9}$$

[Note that $(9)$ would also hold for rational $b$ as long as $kb$ is integer.]