did you learn alternate segment theorem?
solving the question will be easier if you've learnt it, otherwise it's still doable but longer

Need to make use of a few circle properties:
1) QOT is an isos triangle as OQ = OT
2) tangent to the circle at Q is perpendicular to the radius OQ
3) angle QOT is 2x QST as both are subtended by the arc QT

And:
a) Sum of angles in triangle is 180degr
b) (If you haven't learnt alternate segment theorem) Sum of angles in a quadrilateral (SQOT) is 360degr

a = 90 - 62 degr = 28 degr (tan perpendicular to rad)
if you know alternate segment theorem, can use it to find QTS and then subtract a to find x
--
Otherwise
2b = 180 - 2(28) degr = 124 degr (sum of angles in tri)
b = 62 degr (angle at ctr = 2 angle at circumf)
c = 90 - 50 degr = 40 degr (tan perpendicular to rad)
x = 360 - d - c - b
= 2b - c - b
= b - c
= 62 - 40 degr
= 22 degr