L1 = {xn ym zm / m > 0, n > 0 } . Here in L1, the number of y and z are equal.
Option A.
L2 = {xn yn z2m / n > 0, m > 0 } . Here the number of x and y are equal.
L1 $\cap$ L2 = {$x^{n} y^{n} z^{n}$ / n > 0}. This is context sensitive language, not CFL. As L1 is saying the number of y and z are equal and L2 is saying that the number of x and y are equal. Hence All of them are equal.
Option B.
L2 = {xn ym zp / n > m and m > p }. Here the number of y is more than z. But in L1 all the string have the number of y and z equal. So their intersection is $\Phi$. This is a regular language. Hence CFL too.
Option C.
L2 = {xn yn zn / n > 0}. Here L2 is not context free. In question it is asking for CFL.
Only, option A is correct.