Solvent effects on acidity of ethanoic acid
Ethanoic acid has $\mathrm{pK_a}$ values of 4.76, 9.63 and 12.3 in water, methanol and DMSO respectively (source here and here). Comparing this with the $\mathrm{pK_a}$ values given in the question we can see that ethanoic acid is most acidic in water and least acidic in DMSO. $$\ce{pK_{a}(H2O) &lt; pK_{a}(CH3OH) &lt; pK_{a}(DMSO)}$$

In general the acidity of a species in different solvents is affected by three factors: 


The dielectric constant of the solvent. Solvents with larger dielectric constants favour charge separation because the force pulling the charges together as they are separated is lower. This is evidenced by the Born equation which estimates the Gibbs free energy of solvation of gaseous ions: $$\Delta G = -\frac{N_Az^2e^2}{8\pi\epsilon_0r_0}\Bigg(1-\frac{1}{\epsilon_r}\Bigg)$$
The strength of solvation of the conjugate base of the acid.
The strength of solvation of the conjugate acid of the solvent.


Dielectric constant: The dielectric constants of water, methanol and DMSO are 80.1, 32,7 and 46.7 respectively (source). This gives some support to the greater acidity in water but contradicts the observed acidity in methanol and DMSO. This suggests that dielectric constant may not be the major factor that determines the relatives acidities.

Anion solvation: As you suggested in the question, water and methanol are protic solvents but DMSO is aprotic and so the ethanoate ion will be more strongly solvated in water and methanol, shifting the acid-dissociation equilibrium in favour of dissociation. This is considered to be the main factor which explains the greater acidity of ethanoic acid in protic solvents as opposed to aprotic solvents. When comparing water and methanol the strength of anion solvation will be broadly similar because both form $\ce{O-H\bond{~}Anion}$ hydrogen bonds.

Cation solvation: The difference in the stability of the conjugate acid of the solvent may be the factor that causes the difference between water and methanol. The $\mathrm{pK_a}$ of $\ce{CH3OH2+}$ is approximately -2.5 (several sources, none giving exact answers) and the $\mathrm{pK_a}$ of $\ce{H3O+}$ is -1.7. Therefore protonated methanol is less stable than protonated water and so this disfavours dissociation of the acid in methanol. This page claims that this can be attributed to oxygen $p$-orbital donation into $\ce{C-H}~\sigma^*$ orbitals but I don't have enough knowledge of MO theory to verify this (maybe someone else can).

Overall the greater acidity of ethanoic acid in water than methanol seems to be down to the greater dielectric constant of water and the greater stability of the solvent cation formed.

This page says some good things about the subject (some of my ideas were developed from reading this).

Additional reading: This paper looks to have some detail on the anion solvation aspect but I can't access the full text so I'll leave it as a link.