Products of ideals is an ideal and comaximal ideals 
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*Let $I$ and $J$ be two ideals in a ring $R$. Prove that $IJ$ is an ideal. 

*Prove that if $R$ is a commutative ring with two ideals satisfying $I+J=R$ then $IJ=I\cap J$.
I could prove that $IJ$ has an identity, inverse as viewed over addition. I have also prived left and right multiplication property but I am unable to proof the closure over addition.
 A: If $K$ is a nonempty subset of the ring $R$, then $K$ is an ideal of $R$ if and only if the following two conditions are satisfied:
I.)  $K$ is a subgroup of $R$ under the additive ring operation $+$; and
II.)  For any $k \in K$ and $r \in R$, we have $rk, kr \in K$.
Now, in order to establish (I), it in fact suffices to show that for $h, k \in K$, we also have $h - k \in K$; this indeed is a consequence of the following general
Property of Groups:  Let $G$ be a group, and $H$ a nonempty subset of $G$; then $H$ is in fact a subgroup of $G$ if and only if, for $h, k \in H$, $hk^{-1} \in H$.
Proof of Property of Groups:  For if $H$ is a subgroup, it is clear that $h, k \in H$ implies $hk^{-1} \in H$; likewise, if $hk^{-1} \in H$ for $h, k \in H$, the picking any $h \in H$ shows the group identity $e = hh^{-1} \in H$, and hence also $h^{-1} = eh ^{-1} \in H$, and finally, since for $h, k \in H$ we have $k^{-1} \in H$, it follows that $hk = h(k^{-1})^{-1} \in H$, so that in fact $H$ is multiplicatively closed.  End:  Property of Groups.
If we apply the above Property of Groups to the subset $K$ of the ring $R$, we see that to establish (I) we need to show that $k_1 - k_2 \in K$ for all $k_1, k_2 \in K$.  I like this approach because it often involves less work that separately showing that $0 \in K$, $k \in K \Rightarrow -k \in K$, and $k_1, k_2 \in K \Rightarrow k_1 + k_2 \in K$.  I also find it illuminating to see how such techniques fit into a larger context, and after all, we are talking about $(K, +)$ as a (commutative) group.  Hence my little detour through the above Property of Groups.
In order to make further progress, it behooves us to recall the definition $IJ$, as recommended by universalset and DonAntonio in their comments:
$IJ = \{ \sum_l i_l j_l \mid i_l \in I, j_l \in J \}; \tag{1}$
that is, $IJ$ is the set of all sums of the form $\sum_l i_l j_l$, where $i_l \in I$ and $j_l \in J$.  With this definition, it is easy to see that $\sum_l (-i_l) j_l \in IJ$ if and only if $\sum_l i_l j_l$ is, since $-i_l \in I$ for all $i_l \in I$.  Thus for $\sum_l i_l j_l, \sum_k i_k j_k \in IJ$ we have 
$\sum_l i_l j_l - \sum_k i_k j_k = \sum_l i_l j_l + \sum_k (-i_k) j_k \in IJ, \tag{2}$
which shows that $IJ$ satisfies the hypothesis of the Property of Groups with respect to the operation $+$, and is hence an Abelian subgroup of $R$, establishing (I).
As noted by the other writers on this post, showing $IJ$ satisfies (II) is even easier:  for $r \in R$, $r i_l \in I$ and $j_l r \in J$, whence $r \sum_l (i_l j_l) = \sum (r i_l) j_l \in IJ$ and $(\sum_l i_l j_l)r = \sum_l i_l (j_l r) \in IJ$.  
Thus $IJ$ is an ideal in $R$.  That takes care of item (1).
As for item (2), note that whether $R$ is commutative or not, we have $IJ \subset I \cap J$, since any summand of $\sum_l i_l j_l$ lies both in $I$ and $J$, ideals that they are.  So it remains to show that $I \cap J \subset IJ$ under the stated assumptions on $R$.  But as pointed out by YACP, $I + J = R$ implies there are $i \in I, j \in J$ with $i + j =  1$.  Then for $p \in I \cap J$, $p = pi + pj$.  $pi \in IJ$ since $i \in I$ and $p\in I \cap J \subset J$, and similarly $j \in J$ and $p \in I$ yields $pj \in IJ$.  Thus $p = pi + pj \in IJ$ and $I \cap J \subset IJ$.  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: Hint (for 2). $1=a+b$ $\Rightarrow$ $x=ax+bx$.
