Let $I$ and $J$ be ideals of $R$.
a) Prove that $I+J$ is the smallest ideal of R containing both $I$ and $J$.
Proof: First we will show that $I+J$ is an ideal. Let $r \in R$, then $r(I+J)=rI+rJ=I+J$ and thus $I+J$ is an ideal.
Let $K$ be an ideal containing both $I$ and $J$. Then $i+j \in K$ $\forall i \in I,j \in J$. Therefore $I+J$ is the smallest ideal of R containing both $I$ and $J$.
b) Prove that $IJ$ is an ideal contained in $I \cap J$.
First we will show that $IJ$ is an ideal. Let $r \in R$, then $rIJ=(rI)J=IJ$. Similarly $IJr=I(Jr)=IJ$. Thus $IJ$ is an ideal.
Now let $x \in IJ$, wwts $x \in I \cap J$
$x=ij$ for $i \in I, j \in J$
Since I and J are ideals, $ij \in J$ and $ij \in I$. Thus $x \in I \cap J$.
c) Give an example of where $IJ \neq I \cap J$
I need help with this one.
d) Show that if $R$ is a commutative ring and if $I+J=R$ then $IJ=I \cap J$
$IJ \subset I \cap J$ by (b). Now I need to try to show that $I \cap J \subset IJ$.
I have not made much progress in this direction, I need some help. Here are some things that I wrote down.
Let $x \in I \cap J$, wwts that $x=ij$ for $i \in I$ and $j \in J$, where $ij=ji$. Also it may be worth noting that $x=i'+j'$ for some $i' \in I$ and $j' \in J$. So i need to show that given $i'+j'$ there is an $i$ and $j$ such that $i'+j' = ij = ji$.
Alas, let us not forget that $i'+j' \in I$ and $i'+j' \in J$ because $i' + j' \in I \cap J$....
Okay,, now i'm rambling. Are my thoughts for (a) and (b) correct? I wouldn't mind somebody just giving me an appropriate counter example for (c), and then yeah, obviously I need some help with (d) still, I hope my thoughts there are going in the correct direction.
If you want to be formal about showing that $I + J$ and $IJ$ are ideals, you should also show that they are additive subgroups of $R$.
For part (b), as @rschwieb mentioned in the comments, an element of $IJ$ should be a finite sum of elements of the form $ij$ for $i \in I, j \in J$. Your reasoning is still valid though since each term is of the form $ij$ and ideals are closed under addition.
For part (c), consider $R = \mathbb{Q}[x]$ with $I = J = (x)$. Then $IJ = (x^2)$ and $I\cap J = (x)$. Since $x^2$ does not divide $x$, we have that $I \cap J \not \subseteq IJ$.
For part (d), if you are assuming that the given ring has unity, you can write $1 = i + j$ for some $i \in I,j \in J$. So, for $x \in I \cap J$, you can write $x = xi + xj$. Since each term is in $IJ$, we have that $x \in IJ$.
Note: If you aren't assuming that the given ring has unity, it's not necessarily true that $rI = I$ as you used in part (a). Consider the ring $R = 2\mathbb{Z}$. If $I = (4)$ and $r = 2$, then $rI$ does not contain $4$ and so $rI \neq I$.