What is the heat of reaction of weak acid reacting with weak base compared to strong acid reacting with strong base?
Strong acid and strong base

First, lets see where the enthalpy of $\pu{-57.3 kJ/mol}$ comes from. In aqueous solution, strong acids and bases react completely to form $\ce{H+}$ and $\ce{OH-}$, respectively. The neutralization of a strong acid with a strong base then is always the same reaction:

$$\ce{H+(aq) + OH-(aq) -&gt; H2O(l)}$$

We can calculate the enthalpy of reaction from the enthalpies of formation at a temperature of $\pu{25 ^\circ C}$. 

$$
\begin{align}
\Delta H_\mathrm{f}(\ce{H+(aq)}) &amp;= 0 \\
\Delta H_\mathrm{f}(\ce{OH-(aq)}) &amp;= \pu{−230.0 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{H2O(l)}) &amp;= \pu{−285.8 kJ/mol}
\end{align}
$$

Comparing enthalpies of formation for the products with those of the reactants, we get  reaction enthalpy of $\pu{−55.8 kJ/mol}$. The value given in the question is slightly different (maybe for a different temperature, which is not given).

Weak acid and strong base

For this case, the reaction is the following (with $\ce{AH}$ as the weak acid):

$$\ce{AH(aq) + OH-(aq) -&gt; A-(aq) + H2O(l)}$$

The reaction enthalpy for this reaction depends on the nature of the weak acid. Specifically, the answer depends on the difference in the enthalpy of formation of $\ce{A-(aq)}$ vs $\ce{AH(aq)}$. Conceptually, it takes energy to break the $\ce{A-H}$ bond, and the interaction with the solvent water changes when you go from $\ce{AH(aq)}$ to $\ce{A-(aq)}$. Here are two examples:

$$
\begin{align}
\Delta H_\mathrm{f}(\ce{CH3COOH(aq)}) &amp;= \pu{−484.3 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{CH3COO-(aq)}) &amp;= \pu{−486.0 kJ/mol}
\end{align}
$$

Difference is $\pu{-1.7 kJ/mol}$.

$$
\begin{align}
\Delta H_\mathrm{f}(\ce{NH4+(aq)}) &amp;= \pu{−132.5 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{NH3(aq)}) &amp;= \pu{-80.893 kJ/mol}
\end{align}
$$

Difference is $\pu{51.6 kJ/mol}$.

So sometimes the enthalpy is higher, sometimes lower than strong acid/strong base neutralization.

Weak acid and weak base

The weak acid/strong base consideration showed us that once you have a weak acid, the enthalpy depends on the nature of the acid. So can we make a general statement about the enthalpy of a reaction between a weak acid and a weak base? We start by writing down the chemical equation (weak base shown as $\ce{B}$):

$$\ce{AH(aq) + B(aq) &lt;=&gt; A-(aq) + BH+(aq)}$$

The enthalpy of this reaction could be positive or negative (if you read it backwards like this:

$$\ce{BH+(aq) + A-(aq) &lt;=&gt; AH(aq) + B(aq)},$$

you can call $\ce{BH+(aq)}$ the weak acid and $\ce{A-(aq)}$ the weak base, and the sign of the reaction enthalpy switches.

Notice that this is written as equilibrium. Depending on the $\mathrm{p}K_\mathrm{a}$ values of $\ce{AH}$ and $\ce{BH+}$, this reaction will have its equilibrium more on the left or on the right. If the reaction does not go to completion, the enthalpy (based on amount of starting materials) will be less.

Let's take the neutralization of acetic acid and ammonia as a specific example:

$$\ce{CH3COOH(aq) + NH3(aq) &lt;=&gt; CH3COO-(aq) + NH4+(aq)}$$

The $\mathrm{p}K_\mathrm{a}$ values are separated by more than 4 units, so this reaction goes almost to completion. The reaction enthalpy is:

$$
\begin{align}
\Delta H_\mathrm{r} &amp;= \Delta H_\mathrm{f}(\ce{CH3COO-(aq)}) + \Delta H_\mathrm{f}(\ce{NH4+(aq)}) - \Delta H_\mathrm{f}(\ce{CH3COOH(aq)}) - \Delta H_\mathrm{f}(\ce{NH4+(aq)}) \\
  &amp;= \pu{−486.0 kJ/mol} - \pu{132.5 kJ/mol} + \pu{484.3 kJ/mol} + \pu{80.893 kJ/mol} \\
  &amp;= \pu{-53.3 kJ/mol}
\end{align}
$$

That is pretty close to the strong acid/strong base case. 

Back to the question

To prove that 


  "weak acid and weak base reaction will produce lesser heat than constant enthalpy" of strong acid and strong base neutralization 


you would have to do this calculation for every weak acid and weak base combination out there. I would be careful making a statement like this. I think the idea was that because weak acids and bases dissociate much less, less heat is given off (per amount of starting material). However, when you combine a weak acid and a weak base, sometimes the reaction goes almost to completion. As you can see from the example I gave, there are cases for which the enthalpy is very close to that of the strong acid/strong base combination.

Gibbs energy of reaction

For the Gibbs energy of reaction, the situation is much simpler. Weak and strong are defined by the magnitude of acid dissociation equilibrium constants ($K_\mathrm{a}$, and related $\mathrm{p}K_\mathrm{a}$). The difference in $\mathrm{p}K_\mathrm{a}$ of the acid and the conjugate acid of the base is a direct measure of the standard Gibbs energy of reaction: the larger the difference, the larger the Gibbs energy (with appropriate signs, so the strong acid/strong base combination would have the most negative Gibbs energy of reaction, the strong acid/weak base or weak acid/strong base combination would also have negative values, and for the weak acid/weak base combination if would depend on the specifics).