The area in 1) is not bounded by the three relations. So it cannot be calculated, and is in any case not equal to $J(b)$.
Edit: Now that the conditions in (1) has been corrected, it is possible to determine whether or not it is true. And it is, given some assumption like the one in my comment below, namely that if $f(0) = c > 0$, then we set $f^{-1}(y) = 0$ for all non-negative $y. End edit
If $b = f(a)$, then $I(a)$ and $J(b)$ will be the area of two parts of the rectangle with corners $(0, 0), (a, 0), (a, b), (b, 0)$, so in that case $I(a) + J(b) = ab$.
(2) is false, and (3) is true, since if $b\neq f(a)$, then $ab$ will be smaller than the areas of the two integrals. You really should draw a picture of the whole setting, just to convince yourself of this.
A written reasoning for the above paragraph can be done as follows: Assume $b, as is the hypothesis for (2). (3) follow very similar reasoning. Then the rectangle with sides parallel to the axes and with diagonal from the origin to $(a, b)$ will be divided into two by the graph of $f$. The area above the graph is $J(b)$, but the area below will be smaller than $I(a)$, since the contribution to $\int_0^af(x)dx$ from all $x$-values larger than $f^{-1}(b)$ will be limited to height $b$, which is smaller than the height of the graph. So we have $I(a) + J(b) = ab + \int_{f^{-1}(b)}^a(f(x) - b)dx > ab$