`1.`

`a.`

Với `0 < alpha < pi/2` thì `cos alpha >0`

Ta có `cos alpha = sqrt(1 - sin^2 alpha) = sqrt(1 - (3/5)^2) = sqrt(16/25) = 4/5`

`sin (alpha + pi/4) = sin alpha * cos pi/4 + cos alpha * sin pi/4 = 3/5 * sqrt2/2 + 4/5 * sqrt2/2 = (7 sqrt2)/10`

`b.`

Với `pi/2 < alpha < pi` thì `sin alpha >0`

Ta có `sin alpha = sqrt(1 - cos^2 a) = sqrt(1 - ( -4/5)^2) = sqrt(9/25) = 3/5`

`cos (alpha - pi/3) = cos alpha * cos pi/3 + sin alpha * sin pi/3 = (-4/5) * 1/2 + 3/5 * sqrt3/2 = (-4 + 3 sqrt3)/10`

`c.`

Với `pi < alpha < (3pi)/2` thì `cos alpha < 0`

Ta có `cos alpha = - sqrt( 1 - sin^2 alpha ) = - sqrt( 1 - ( -3/5)^2) = - sqrt(16/25) = - 4/5`

`cos (alpha + pi/6) = cos alpha * cos pi/6 - sin alpha * sin pi/6 = (-4/5) * sqrt3/2 - ( -3/5) * 1/2 = (-4 sqrt(3) + 3)/10`