Assuming the question I linked is the same as yours, you should have the following information (provided you completed exercise 12):
[tex]\mu_X=1.01[/tex]
[tex]\mu_Y=1.23[/tex]
[tex]\sigma_X\approx0.89994[/tex]
[tex]\sigma_Y\approx0.99855[/tex]
[tex]\mathrm{Cov}(X,Y)\approx0.2377[/tex]
[tex]\rho_{X,Y}\approx0.264512[/tex]
Given that [tex]Z=X+Y[/tex], we have
[tex]\mu_Z=\mu_X+\mu_Y=2.24[/tex]
[tex]\sigma_Z=\sigma_X+\sigma_Y+2\,\mathrm{Cov}(X,Y)\approx1.51076[/tex]
Finally, [tex]Z=2[/tex] is the same as the event that any one of the following occurs: [tex]X=0[/tex] and [tex]Y=2[/tex]; [tex]X=1[/tex] and [tex]Y=1[/tex]; or [tex]X=2[/tex] and [tex]Y=0[/tex]. So we have
[tex]\mathbb P(Z=2)=0.02+0.16+0.07=0.25[/tex]
