When an independent voltage source is expressed as: 50e−tV What does that mean?
As mentioned by pjc50 already, \$e^{-t}\$ means the voltage drops with time like the inverse logarithm. This means that at time \$t=0\$, the input voltage is 50 (because \$50 \cdot e^{-0} = 50\$), at \$t=1\$ the value is \$\frac{50}{e}\$, and so on. 

Just algebra will not get you an answer here, you need to integrate. In fact, the answers are quite straightforward:

\$E = \int_{t=0}^{t=250ms}{p(t)dt}\\
Q = \int_{t=0}^{t=250ms}{i(t)dt}\$

Of course, the actual evaluation of these integrals is the hard part. Because you seem to be having some trouble with that as well, let's quickly try that as well. The great thing about the e-curve is that \$\frac{d}{dt}e^x = e^x\$, no matter what you use for \$x\$. So the integral of \$e^x\$ is also simply \$e^x\$. If this is not immediately clear you should practice your calculus.

So this gives us enough information to solve the energy integral. K is the integration constant.

\$\begin{eqnarray}E &amp;=&amp; \int_{t=0}^{t=250ms}{p(t)dt}\\
&amp;=&amp;\int_{t=0}^{t=250ms}{2.5e^{-4t}dt}\\
&amp;=&amp;[-\frac{5}{8}e^{-4t}]_{0}^{0.25} + K\\
&amp;=&amp;-\frac{\frac{5}{8}}{e} + \frac{5}{8} + K\\
&amp;=&amp;0.39507534926 + K\end{eqnarray}\$

We know that K=0 because the integral has a value of 0 when integrating an infinitesimally small amount of time starting at \$t=0\$, so the end result is \$E=0.395 \hspace{3pt} J\$.

For Q, you need to know the current. Obviously, because of ohm's law \$P=IV\$, so you need to calculate:

\$\begin{eqnarray}Q &amp;=&amp; \int_{t=0}^{t=250ms}{i(t)dt}\\
&amp;=&amp; \int_{t=0}^{t=250ms}{\frac{p(t)}{v(t)}dt}\\
&amp;=&amp; \int_{t=0}^{t=250ms}{\frac{2.5e^{-4t}}{50e^{-t}}dt}\\
&amp;=&amp; \int_{t=0}^{t=250ms}{\frac{2.5}{50}e^{-3t}dt}
 \end{eqnarray}\$

And I leave it to you to actually calculate that integral, it should be very easy given the examples in this answer.