1) You need to solve exponential equation `4^(x-1) - 3*2^(x-2) = 1` , hence, you should come up with the substitution `2^x = y`  such that:

`4^x/4 - 3*2^x/4 = 1`

`(2^(2x)) - 3*2^x - 4 = 0`

`y^2 - 3y - 4 = 0`

You need to use quadratic formula such that:

`y_(1,2) = (3+-sqrt(9+16))/2`

`y_1 = (3+5)/2 =gt y_1 = 4`

`y_2 = (3-5)/2 =gt y_2 = -1`

You need to solve for x the equations `2^x = y_1 =gt 2^x = 4 =gt 2^x = 2^2 =gt x = 2`

`2^x = y_2 =gt 2^x = -1`  contradiction since `2^x gt 0` .

Hence, the solution to equation is `x = 2` .

2) You need to expand `tan(a - pi/4)`  such that:

`tan(a - pi/4) = (tan a - tan(pi/4))/(1 + tan a*tan (pi/4))`

`tan(a - pi/4) = (tan a - 1)/(1 + tan a*1)`

The problem provides the information that `tan(a - pi/4) = 3/4` , such that:

`(tan a - 1)/(1 + tan a*1) = 3/4`

`4(tan a - 1) = 3(1 + tan a)`

You need to open the brackets such that:

`4tan a - 4 = 3 + 3tan a`

You need to subtract `3 tan a`  such that:

`4 tan a - 3tan a = 3 + 4`

`tan a = 7`

Hence, evaluating `tan a`  under given conditions yields `tan a = 7.`