Let $g(x) = \frac{f(x)+f(-x)}{2}$, and $h(x) = \frac{f(x)-f(-x)}{2}$.
Then $f(x) = g(x) + h(x)$, $g(x)$ is an even function, $h(x)$ is an odd function, and $g(x+2\pi) = g(x)$, $h(x+2\pi) = h(x)$ for any $x \in \mathbb{R}$.

Define
$$f_1(x) = \frac{g(x) + g(x + \pi)}{2},$$
$$f_2(x) = \begin{cases} \frac{g(x) - g(x + \pi)}{2\cos x}, & x \neq k\pi + \frac{\pi}{2}, \\ 0, & x = k\pi + \frac{\pi}{2}, \end{cases}$$
$$f_3(x) = \begin{cases} \frac{h(x) - h(x + \pi)}{2\sin x}, & x \neq k\pi, \\ 0, & x = k\pi, \end{cases}$$
$$f_4(x) = \begin{cases} \frac{h(x) + h(x + \pi)}{2\sin 2x}, & x \neq \frac{k\pi}{2}, \\ 0, & x = \frac{k\pi}{2}, \end{cases}$$
where $k$ is an arbitrary integer. It is easy to check that $f_i(x)$ ($i = 1, 2, 3, 4$) satisfy (1).

Next we prove that $f_1(x) + f_2(x)\cos x = g(x)$ for any $x \in \mathbb{R}$. When $x \neq k\pi + \frac{\pi}{2}$, it is obviously true. When $x = k\pi + \frac{\pi}{2}$, we have
$$f_1(x) + f_2(x) \cos x = f_1(x) = \frac{g(x) + g(x + \pi)}{2},$$
and
$$\begin{aligned} g(x + \pi) &= g\left(k\pi + \frac{3\pi}{2}\right) = g\left(k\pi + \frac{3\pi}{2} - 2(k+1)\pi\right) \\ &= g\left(-k\pi - \frac{\pi}{2}\right) = g\left(k\pi + \frac{\pi}{2}\right) = g(x). \end{aligned}$$
The proof is complete.

Further we prove that $f_3(x)\sin x + f_4(x)\sin 2x = h(x)$ for any $x \in \mathbb{R}$. When $x \neq \frac{k\pi}{2}$, it is obviously true. When $x = k\pi$, we have
$$h(x) = h(k\pi) = h(k\pi - 2k\pi) = h(-k\pi) = -h(k\pi).$$
That means $h(x) = h(k\pi) = 0$. In this case,
$$f_3(x)\sin x + f_4(x)\sin 2x = 0.$$
$$\text{Therefore } h(x) = f_3(x)\sin x + f_4(x)\sin 2x.$$
When $x = k\pi + \frac{\pi}{2}$, we have
$$\begin{aligned} h(x + \pi) &= h\left(k\pi + \frac{3\pi}{2}\right) = h\left(k\pi + \frac{3\pi}{2} - 2(k+1)\pi\right) \\ &= h\left(-k\pi - \frac{\pi}{2}\right) = -h\left(k\pi + \frac{\pi}{2}\right) = -h(x). \end{aligned}$$
So
$$f_3(x)\sin x = \frac{h(x) - h(x + \pi)}{2} = h(x).$$
Furthermore, $f_4(x)\sin 2x = 0$. Therefore
$$h(x) = f_3(x)\sin x + f_4(x)\sin 2x.$$
This completes the proof.

In conclusion, $f_i(x)$ ($i = 1, 2, 3, 4$) satisfy (2).