Let `P(E_(2))=x`.
Then, `E_(1)` and `E_(2)`being independent events, we have
`P(E_(1) nn E_(2))=P(E_(1))xxP(E_(2))=0.35xx x=0.35 x`.
Now, `P(E_(1) uu E_(2))=P(E_(1))+P(E_(2))-P(E_(1) nn E_(2))`
`implies 0.60=0.35+x-0.35 x`
`implies 0.65 x=0.25`
`implies x=0.25/0.65=25/65=5/13`.
Hence, `P(E_(2))=5/13`.