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y^2 - y -2x -1 = 0

Use implicit differentiation

2y*y' - 1*y' - 2 = 0

=> y'(2y - 1) = 2

=> y' = 2/(2y - 1)

The tangent to a curve at a point has a slope equal to the value of the first derivative at the point.

At (-2, 1), y' = 2/(1)

The line through (-2, 1) with a slope 2 is (y - 1)/(x + 2) = 2

=> y - 1 = 2x + 4

=> 2x - y + 5 = 0

The slope of the normal at the point is -1/2

The equation of the normal is (y - 1)/(x + 2) = -1/2

=> 2(y - 1) = -x - 2

=> 2y + x = 0

**The equation of the tangent at the given point is 2x - y + 5 = 0 and of the normal is 2y + x = 0**