Matrice combination

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is the product of translation matrices equal to the combination of the translation transformations?

Yes. The composition of two translations (in ordinary 3d space) is a new translation and it is the product of the two translations.
Let

     [1  0  0  dx1]       [1  0  0  dx2]T1 = [0  1  0  dy1], T2 = [0  1  0  dy2]     [0  0  1  dz1]       [0  0  1  dz2]     [0  0  0   1 ]       [0  0  0   1 ]

be two translations. Then

                    [1  0  0  dx1+dx2]T = T2*T1 = T1*T2 = [0  1  0  dy1+dy2]                    [0  0  1  dz1+dz2]                    [0  0  0     1   ]

which is a new translation. This is be generalized for arbitrary number of translations.

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Is it correct to say that they are commutive transformations?

Yes. Simply notice that T1*T2 = T2*T1, for any two translations.

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Are they also associative?

Translation matrices are associative under multiplication, because all matrices are associative under multiplication.

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Can the same be said about rotation matrices?

The composition of any two rotations is another rotation. This equivalent to saying that the set of all rotations (in 3d space) is closed under matrix multiplication. Also, the composition of any two reflections is a rotation.
As for commutation, any two matrices commute if-f they have the same eigenvalues. For rotation matrices, this means that -in order to commute- they must rotate around the very same axis. (in general)
Therefore two rotation matrices commute if they operate on the same plane.
It also holds that infitesimal rotations commute, but this is another story. It has to do with allowing for some error in the process.
And again, they are associative under composition (multiplication essentially) because all matrices are.

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But the product of translation and rotation matrices are neither commutivite nor associative? Is that correct?

Such mixed products are generally non-commutative.