a) Solve the differential equation: (x+1)dydx−3y=(x+1)4 given that y=16 and x=1, expressing the answer in...

Good job on solving part (a) of the question. That was the tough part. Part (b) is much more easier.
As you stated, we get that the function is:
${\displaystyle y={(1+x)}^4}$
Now, in order to find the area between this function, the function ${\displaystyle f\left(x\right)={(1-x)}^4}$, and the x-axis, we need to the find intersection points:
${\displaystyle {(1+x)}^4={(1-x)}^4}$
${\displaystyle {\left({(1+x)}^4\right)}^{\frac{1}{4}}={\left({(1-x)}^4\right)}^{\frac{1}{4}}}$
$1+x=1-x$
${\displaystyle x=0}$
When ${\displaystyle x=0}$, ${\displaystyle {(0+1)}^4={(0-1)}^4=1}$.Thus, we get that the two functions intersect at the point (0,1).
Next, we look at where each of the two functions intersect the x-axis. We get that:
${\displaystyle {(1+x)}^4=0}$
${\displaystyle 1+x=0}$
${\displaystyle x=-1}$
${\displaystyle {(1-x)}^4=0}$
${\displaystyle 1-x=0}$
${\displaystyle x=1}$
Therefore, we get our two endpoints: (-1,0) and (1,0).
Thus, we can use construct our integral now:
${\displaystyle A={\int }_{-1}^0{(1+x)}^{4}dx+{\int }_0^1{(1-x)}^{4}dx}$
${\displaystyle {\left[\frac{{(x+1)}^5}{5}\right]}_{\{-1\}}^0+{\left[\frac{{(x-1)}^5}{5}\right]}_0^1=\frac{1}{5}-0+0-(-\frac{1}{5})=\frac{2}{5}}$
Hope that helped. Comment if you have any questions.