Let $C_i\sim C$ and $R_i \sim R$ denote the durations (in ticks) of the $i$th coupled periods of clear and rainy weathers, respectively. We want to find
$$p(t)=\mathbb P \left (A_t:=\text{at time $t$ the weather is $\bf{rainy}$} \right ).$$
Let us define
$$q(t)=1- p(t)=\mathbb P \left (A'_t:=\text{at time $t$ the weather is $\bf{clear}$} \right ).$$
By conditioning on $C_1$, we have
$$q(t)=\mathbb P (A'_t \, |C_1>t)\mathbb P (C_1>t)+\mathbb P (A'_t, C_1 \le t).$$
By observing
$$\mathbb P (A'_t \, |C_1>t)=1$$
$$\mathbb P (A'_t, C_1 \le t)=\int_{0}^{\infty} \mathbb P (A'_t, C_1 \le t|R_1+C_1=x) \text{d}F_{R_1+C_1}(x)= \int_{0}^{t} q(t-x) \text{d}F_{R_1+C_1}(x),$$
we obtain
$$\color{blue}{p(t)=1- q(t) \\ q(t)=1-F_{C}(t)+\int_{0}^{t} q(t-x) \text{d}F_{R+C}(x).}$$
When $F_{R+C}$ has a pdf, denoted by $f_{R+C}$, it is a linear Volterra integral equation, which can be solved by Laplace transformation. The final solution is
$$\color{blue}{p(t)= 1-\mathcal L^{-1} \left [\frac{\mathcal L[1-F_{C}](s)}{1-\mathcal L[f_{R+C}](s)} \right ](t)}.$$
For $C \sim \alpha+U(0,a)$ and $R\sim \alpha + U(0,b)$ with $\alpha=12000, a=168000 , b=12000$, from the above you can find $p(t)$ by considering that ($a>b$)
$$F_{C}(t)=\min \left(1,\frac{t-\alpha}{a} \right )I_{(\alpha,\infty)}(t)$$
$$f_{R+C}(x)=\frac{1}{ab}(x-2\alpha)I_{(2\alpha,2\alpha+b)}(x)+\frac{1}{a}I_{(2\alpha+b,2\alpha+a)}(x)+\left (\frac{1}{a}-\frac{1}{ab}(x-2\alpha-a) \right )I_{(2\alpha +a,2\alpha + a + b)}(x).$$
After determining $\mathcal L[1-F_{C}]$ and $\mathcal L[f_{R+C}](s)$, see here and here, the above gives (for presentation simplicity to obtain the following, the time unit is in thousands of ticks):
$$1-\mathcal L^{-1} \left [\frac{\frac{e^{-180 s} - e^{-12 s} + 168 s}{168 s^2}}{1 - \frac{e^{-204 s} (1 - e^{12 s} - e^{168 s} + e^{180 s})}{2016 s^2}} \right ](t).$$
The inner of the inverse Laplace transform can be simplified as
$$\frac{12 e^{24 s} (1 - e^{168 s} + 168 e^{180 s} s)}{ e^{12 s} + e^{168 s} - e^{180 s} + 2016 e^{204 s} s^2-1}.$$
After checking many options, it seems there is no closed-form expression for this function. WolframAlpha could not obtain the inverse Laplace transform, see here. Also, Mathematica could not find any closed-form solution.
Finally, I asked some experts to help in this Wolfram question. Fortunately, we can find the inverse numerically. I used the final result obtained by Mariusz Iwaniuk based on the GWR method in a reply to my question among different methods, which seems to be more reliable (for more details, codes, etc. see that question). Thus, the final solution for $p(t)$ can be depicted as follows:
Limiting case:
As $q(t)$ is a solution for
$$p(t)=1-F_{C}(t)+\int_{0}^{t} p(t-x) \text{d}F_{R+C}(x),$$
from here, it can be expressed as
$$q(t)=1-F_{C}(t)+\int_{0}^{t} (1- F_{C}(t-x))\text{d}m(x)$$
where $m(t)$ is the mean of the number of renewals by time $t$ in a renewal process in which the distribution of the inter-arrival times follows $R+C$. From the above we can see that
$$\lim_{t \to \infty}q(t)=\lim_{t \to \infty}(1-F_{C}(t))+\lim_{t \to \infty}\int_{0}^{t} (1- F_{C}(t-x))\text{d}m(x)=0+\frac{1}{\mathbb E (R+C)}\int_{0}^{\infty} (1- F_{C}(x))\text{d}x=\frac{\mathbb E (C)}{\mathbb E (R)+\mathbb E (C)}. \tag{2}$$
Thus,
$$\color{blue}{\lim_{t \to \infty}p(t)}=1-\lim_{t \to \infty}q(t)= \color{blue}{\frac{\mathbb E (R)}{\mathbb E (R)+\mathbb E (C)}}=\frac{\frac{12000+24000}{2}}{\frac{12000+180000}{2}+\frac{12000+24000}{2}}=\color{blue}{\frac{3}{19}}.$$
The last limit in (2) is obtained from the key renewal theorem:
$$\lim_{t \to \infty} \int_{0}^{t} g(t-x)\text{d}m(x)=\frac{1}{\mu}\int_{0}^{\infty} g(x)\text{d}x,$$
which holds for any non-decreasing function $g: \mathbb R_{\ge0} \to \mathbb R_{\ge0}$ with finite $\int_{0}^{\infty} g(x)\text{d}x$ whenever the distribution of the interval arrivals in the renewal process with $m(x)$ is non-lattice and has the finite mean $\mu$.
