Analyzing RC circuit that has two "switch"
Well, for \$t&lt;0\$ we know that the circuit is equivalent to:



simulate this circuit &ndash; Schematic created using CircuitLab
The voltage that is created across the capacitor for \$t&lt;0\$, given by:
$$\text{V}_\text{C}\left(0\right):=\lim_{t\to\infty}\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{sC}}\cdot\frac{\text{R}_1}{\text{R}_1+\text{R}_2+\left(\text{R}_3\left.\right\vert\left.\right\vert\frac{1}{\text{sC}}\right)}\cdot\frac{\text{R}_3}{\text{R}_3+\frac{1}{\text{sC}}}\cdot\frac{\text{I}_x}{\text{s}}\right]_{\left(t\right)}=$$
$$\lim_{\text{s}\to0}\text{s}\cdot\frac{1}{\text{sC}}\cdot\frac{\text{R}_1}{\text{R}_1+\text{R}_2+\left(\text{R}_3\left.\right\vert\left.\right\vert\frac{1}{\text{sC}}\right)}\cdot\frac{\text{R}_3}{\text{R}_3+\frac{1}{\text{sC}}}\cdot\frac{\text{I}_x}{\text{s}}=\frac{\text{I}_x\text{R}_1\text{R}_3}{\text{R}_1+\text{R}_2+\text{R}_3}\tag1$$
Where I used the final value theorem of the Laplace transform and \$\$ means the parallel resistor equation.
Now, for \$0\le t&lt;\tau\$ we have the following equivalent circuit:



simulate this circuit
Using the discharge equation of the capacitor we get (you can also show that using KVL and KCL), we know that:

$$\text{V}_\text{C}\left(t\right)=\text{V}_\text{C}\left(0\right)\cdot\exp\left(-\frac{t}{\text{CR}_3}\right)\tag2$$
$$\text{I}_\text{C}\left(t\right)=\frac{\text{V}_\text{C}\left(0\right)}{\text{R}_3}\cdot\exp\left(-\frac{t}{\text{CR}_3}\right)\tag3$$

So, for \$0\le t&lt;\tau\$ we get that the total power in the resistor \$\text{R}_3\$ is given by:
$$\text{P}_{0\le t&lt;\tau\text{|}\text{R}_3}=\text{R}_3\left(\frac{\text{I}_x\text{R}_1}{\text{R}_1+\text{R}_2+\text{R}_3}\right)^2\cdot\exp\left(-\frac{2\tau}{\text{CR}_3}\right)\tag4$$
So, for the part when \$t\ge\tau\$, we get the same voltage equation as above. But with a different starting voltage and different resistive value:
$$\text{V}_{t\ge\tau\text{|}\text{C}}\left(t\right)=\text{V}_{0\le t&lt;\tau\text{|}\text{C}}\left(\tau\right)\cdot\exp\left(-\frac{t}{\text{C}}\cdot\frac{\text{R}_3+\text{R}_4}{\text{R}_3\text{R}_4}\right)\tag5$$
So, the total power is equal to:

For \$\text{R}_3\$:
$$\text{P}_{\text{total}\text{|}\text{R}_3}=\text{P}_{0\le t&lt;\tau\text{|}\text{R}_3}+\frac{\text{V}_{t\ge\tau\text{|}\text{C}}^2\left(t\right)}{\text{R}_3}\tag6$$
For \$\text{R}_4\$:
$$\text{P}_{\text{total}\text{|}\text{R}_4}=\frac{\text{V}_{t\ge\tau\text{|}\text{C}}^2\left(t\right)}{\text{R}_4}\tag7$$


Using your values I got:

For \$\text{R}_3\$:
$$\text{P}_{\text{total}\text{|}\text{R}_3}=\frac{32}{125\exp\left(\frac{2}{5}\right)}+\frac{32}{125}\cdot\exp\left(-\frac{4}{5}-100t\right)\tag6$$
For \$\text{R}_4\$:
$$\text{P}_{\text{total}\text{|}\text{R}_4}=\frac{8}{125}\cdot\exp\left(-\frac{4}{5}-100t\right)\tag7$$