First, in general $I^a(X;Y) \geq 0$ does not hold. One can find easy counterexamples with just two states.

The other part of inequality (1) does hold.
For inequality (2), the reverse does actually hold. And with that, inequality (3) is trivially true.

We show that $\frac{\partial I^a}{\partial a} \geq 0$ for $a \leq 0$ and $\frac{\partial I^a}{\partial a} \leq 0$ for $a \geq 0$. So $a \mapsto I^a$ is maximal at $a=0$.

I write $p(x) = p_1(x)$ and $p(y) = p_2(y)$ for clarity. It holds
\begin{align}
\frac{\partial I^a(X;Y)}{\partial a} &= \frac{\partial}{\partial a}\Big(p(x_{\bar{i}}, y_{\bar{j}}) \log\frac{p(x_{\bar{i}}, y_{\bar{j}}) + a}{(p_1(x_{\bar{i}})+a)p_2(y_{\bar{j}})} + \sum_{j \neq \bar{j}} p(x_{\bar{i}}, y_j) \log\frac{p(x_{\bar{i}}, y_{j})}{(p_1(x_{\bar{i}})+a)p_2(y_{j})}\Big) \\
&=  \Big(\frac{p(x_{\bar{i}}, y_{\bar{j}})}{p(x_{\bar{i}}, y_{\bar{j}}) + a} - \frac{p(x_{\bar{i}}, y_{\bar{j}})}{p_1(x_{\bar{i}})+ a}\Big) - \sum_{j \neq \bar{j}} \frac{p(x_{\bar{i}}, y_j)}{p_1(x_{\bar{j}})+a} \\
&= \frac{p(x_{\bar{i}}, y_{\bar{j}})}{p(x_{\bar{i}}, y_{\bar{j}})+a} - \frac{p_1(x_{\bar{i}})}{p_1(x_{\bar{i}})+a}
\end{align}
And since $[0, 1] \ni x\mapsto \frac{x}{x+a}$ is increasing for $a \geq 0$ and decreasing for $a \leq 0$, and $p_1(x_{\bar{i}}) \geq p(x_{\bar{i}}, y_{\bar{j}})$, the claim follows.