Peroxide effect in reagents apart from HBr and CHCl₃
Is the peroxide effect observed only with addition of hydrogen bromide
  and chloroform?


No, the "peroxide effect" or "anti-Markovnikov" addition of a free radical to an unsaturated bond is observed with other compounds too, $\ce{CCl4}$ is another  example.

Also, just to be clear, these reactions involve the addition of radicals ($\ce{X^{.}}$), not electrophiles, to alkenes (or alkynes).

Let's examine the radical addition of $\ce{HX}$ to an olefin.  I'll use a general method that you can later use to examine any radical addition to an unsaturated bond.

There are two propagation steps that are key to our analysis:
$$\ce{X* + H2C=CH2 -&gt; X-H2C-CH2. ~~(1)}$$
$$\ce{HX + X-H2C-CH2. -&gt; X-H2C-CH2-H ~~(2)}$$

If two halogen radicals combine, or if a halogen radical and a halo-alkyl radical (formed in the propagation steps) combine, the reaction is terminated. The propagation steps are in competition with the termination steps. If the propagation steps are endothermic, then they compete less effectively with the termination steps and the radical addition will not proceed in a useful manner.

In propagation step (1) we are breaking a pi bond and making a $\ce{C-X}$ bond, let's look at the thermodynamics of this  step.

\begin{array}{|c|c|c|c|} \hline
\ce{C-X} &amp; \text{Bond Strength} &amp; \text{Pi Bond Strength} &amp; \text{Overall}  \\ &amp; \text{(kcal/mol)} &amp; \text{(kcal/mol)} &amp; \text{(kcal/mol)}\\ \hline
\ce{C-F} &amp; 116 &amp; 66 &amp; -50\\ \hline
\ce{C-Cl} &amp; 81 &amp; 66 &amp; -15\\ \hline
\ce{C-Br} &amp; 68 &amp; 66 &amp; -2\\ \hline
\ce{C-I} &amp; 51 &amp; 66 &amp; 15\\ \hline
\end{array}

We see that this step is exothermic for the first 3 hydrogen halides, but is endothermic for the $\ce{HI}$ case.  Already we see that hydrogen iodide will not add to olefins via a radical pathway at room temperature.

In propagation step (2) we are breaking an $\ce{HX}$ bond and making a $\ce{C-H}$ bond.  Let's perform a similar thermodynamic analysis of this step.

\begin{array}{|c|c|c|c|} \hline
\ce{H-X} &amp; \text{Bond Strength} &amp; \ce{C-H}~ \text{Bond Strength} &amp; \text{Overall}  \\ &amp; \text{(kcal/mol)} &amp; \text{(kcal/mol)} &amp; \text{(kcal/mol)}\\ \hline
\ce{H-F} &amp; 135 &amp; 99 &amp; 36\\ \hline
\ce{H-Cl} &amp; 103 &amp; 99 &amp; 4\\ \hline
\ce{H-Br} &amp; 88 &amp; 99 &amp; -11\\ \hline
\ce{H-I} &amp; 71 &amp; 99 &amp; -28\\ \hline
\end{array}

(source for bond strengths)

Both propagation steps (1) and (2) must occur for the reaction to proceed.  Only in the case of the addition of $\ce{HBr}$ is the reaction exothermic for both steps.  For the hydrogen halide series, only $\ce{HBr}$ will add to an olefin by a free radical mechanism.