Let $\gamma:[a,b]\to\mathbb{C}$ be a continuous curve.
1) Is it true that one can find a sequence of numbers $(r_n)_{n\in\mathbb{N}}\subset (0,\infty)$ and some function $\varphi:\mathbb{R}\times [a,b]\to\mathbb{R},\ \varphi\in C^{1}$ such that the sequence of curves:
$$\gamma_n:[a,b]\to\mathbb{C},\ \gamma_{n}(t)=\sum_{k=0}^n r_k e^{i\varphi(k,t)}$$
converges uniformly to $\gamma$ on $[a,b]$? Moreover can we choose $(r_n)_{n\in\mathbb{N}}$ to be decreasing or strictly decreasing? Can we choose $\varphi (k,t)=2k\pi t$?
2) Is it true that there is a function $r:\mathbb{R}\to (0,\infty),\ r\in L^1 (\mathbb{R})$ such that $r(k)=r_k,\ \forall k\in\mathbb{N}$ and
$$\gamma(t)=\int_{-\infty}^{\infty}r(x)e^{i2\pi x t} dx\ ?$$
Edited following the original post's edit
($1$) For the first one, that's still not possible, consider the sawtooth function
$$s(t) = {1\over 2}-t, 0\le t\le 1.$$
Then a uniformly convergent sequence of continuous $L^1(\Bbb T)$ (i.e. periodic) functions is also continuous and periodic. However
$$\lim_{t\to 1-}s(t)=-{1\over 2}\quad \lim_{t\to 1+}s(t)={1\over 2}$$
so that the limiting function cannot be periodic. You may argue we don't start with a periodic function, $s(0)\ne s(1)$, however remember that $L^1$ functions are equal a.e., so that the series has to be the same as the a.e. equal function where we define it only on $[0,1)$ and extend by periodicity. But we know that Fourier series converge to ${1\over 2}(f(x^+)+f(x^-))$ around every point, so the limiting function would have value $0$ at $0$ and left-hand limit ${1\over 2}$, and so cannot be the continuous function, $s(t)$.
For the second part, the counterexample from before is no less true when you change it to "moreover can we" followed by the same hypotheses, i.e. you can still just consider $\gamma(t)\equiv 1$ and let $I$ be whatever you want. The uniqueness of Fourier series says that only the sequence
$$c_n=\begin{cases} 1 & n=0 \\ 0 & n\ne 0\end{cases}$$
converges to the constant function $1$ (even point-wise), among all sequences of linear combinations of the functions $\{e^{2\pi i n}\}_{n\in\Bbb Z}$, but your sequence $r_k$ is $0$ for $k=0$.
($2$) Clearly if this is so then
$$|\gamma(t)|=\left|\int_{\Bbb R}r(x)e^{2\pi i x}\,dx\right|\le \lVert r\rVert_1<\infty$$
so choose any $\gamma$ which is unbounded and that is a counterexample, eg. $\gamma(t)=t$ and $I=\Bbb R$.