I'm getting a length of 4 and a width of 8. I don't know if this matches your 8x4. Usually length is specified first. I'm assuming that length corresponds to the measure from the top to the bottom.

for (b) just use the product rule

$\dfrac d {dt} (2t-5)e^{3t-1} = \left(\dfrac{d}{dt}(2t-5)\right) \cdot e^{3t-1} + (2t-5)\cdot \left(\dfrac{d}{dt}e^{3t-1}\right)$