Electrical – How to find the frequency of the AC source in RL circuit

You won't (easily) get a closed form for \$\int e^{p(t)}dt\$.

The solution to the differential equation that you wrote is:

$$i(t)=\left(\frac{V}{L}\right)e^{\frac{-R_{mod}\sin(\omega t)}{L\omega}-\frac{R_{offset}t}{L}}\left(\int_0^t e^{\frac{R_{mod}\sin(\omega \tau)}{L\omega}+\frac{R_{offset}\tau}{L}} d\tau\right)+ Ce^{\frac{-R_{mod}\sin(\omega t)}{L\omega}-\frac{R_{offset}t}{L}}$$ where \$C\$ is an unknown determined by the initial condition. Obviously the term involving integration is the troublesome bit so lets focus on that piece.

The power series for \$e^{x}\$ converges uniformly on bounded sets so we can exchange the power series with the integral giving $$\int_0^t e^{\frac{R_{mod}\sin(\omega \tau)}{L\omega}+\frac{R_{offset}\tau}{L}} d\tau = \sum_{n=0}^{\infty}\frac{1}{n!}\int_0^t\left(\frac{R_{mod}\sin(\omega \tau)}{L \omega}+\frac{R_{offset} \tau}{L} \right)^n d\tau.$$

Integrating the inner terms involves integrating terms of the form \$\tau^k\sin(\omega \tau)^{n-k}\$ which have closed form solutions.

The real upshot is that if we only want approximate solutions we can bound the tail of this series by bounding the integrands and using that $$\sum_{n=N+1}^{\infty}\frac{1}{n!}x^n = e^x-\frac{e^x \Gamma(N+1,x)}{N!}.$$

Then we will have a closed formed approximate solution \$\tilde{i}_N(t)\$ which approximates \$i(t)\$ and we will know a bound for the error \$|\tilde{i}_N(t) - i(t)|\$.

You have calculated the impedances correctly. Now since you know the voltage across R1 (given to be 40V) you can easily calculate the current in that branch from V=I/R. Since you know the impedance, you can then calculate U in volts RMS.

From U in volts RMS and the impedance of the second branch, you can calculate the current in that branch.

Since you now know both currents, the power is just I^2R in each branch, since the reactances are lossless. No need to consider the angles at all (directly). The total power is just the sum of the powers dissipated in each branch by the respective resistor.

The correct answer is in the listed numbers, so you should be fine.