First part,

`(1+cos(x+2pi))/(1-cos(x))`

we know, any trignometric function is repeated after 2pi value, therefore,

`cos(x) = cos(x+2pi)`

applying this in above expression,

`=(1+cos(x))/(1-cos(x+2pi))`

`=(cos(x))/(cos(x+2pi))((1/cos(x)+1)/(1/cos(x+2pi)-1))`

now this reduces to,

`=(sec(x)+1)/(sec(x+2pi)-1)` Proved.

The second part,

`1/(4sin^2(x)cos^2(x)) - (1 -tan^2(x))/(4tan^2x)`

since 2sin(x)cos(2x) = sin(2x) and `tan(2x) = (2tan(x))/(1-tan^2(x))`

the above expression reduces to,

`=1/(sin^2(2x)) - 1/(tan^2(2x))`

`=1/(sin^2(2x))-(cos^2(2x))/(sin^2(2x)) `

`=(1-cos^2(2x))/(sin^2(2x))`

but we know. `sin^2(2x)+cos^2(2x) = 1`

therefore,

`=(sin^2(2x))/(sin^2(2x))`

`=1`

The answer is 1.