(a) In the diagram. \(\over{Rs}\) and \(\over{RT}\) are tangent to the circle with centre O, < TUS = 68\(^o\), < SRT = x and < UTO = y. Find the value of x....

(a) Finding \(x = \angle SRT\).

From the diagram, \(RT\) and \(RS\) are tangents drawn from the external point \(R\), touching the circle (centre \(O\)) at \(T\) and \(S\). \(\angle TUS = 68^\circ\) is the angle at the circumference standing on chord \(TS\).

The angle subtended at the centre by the same chord \(TS\) is twice the angle at the circumference:

\[ \angle TOS = 2 \times \angle TUS = 2 \times 68^\circ = 136^\circ \]

Since a tangent is perpendicular to the radius at the point of contact:

\[ \angle OTR = \angle OSR = 90^\circ \]

Now consider quadrilateral \(RTOS\). Its interior angles sum to \(360^\circ\):

\[ \angle SRT + \angle OTR + \angle TOS + \angle OSR = 360^\circ \]

\[ x + 90^\circ + 136^\circ + 90^\circ = 360^\circ \]

\[ x = 360^\circ - 316^\circ = 44^\circ \]

(b) Capacity of each tank. Let the capacity of tank \(B\) be \(b\) litres. Then tank \(A\) holds \((b + 600)\) litres.

After pumping out \(100\) litres from each, tank \(A\) has \(3\) times as much as tank \(B\):

\[ (b + 600) - 100 = 3\,(b - 100) \]

\[ b + 500 = 3b - 300 \]

\[ 500 + 300 = 3b - b \]

\[ 800 = 2b \;\Rightarrow\; b = 400 \]

So tank \(B = 400\) litres and tank \(A = 400 + 600 = 1000\) litres.

Check: \(1000 - 100 = 900\) and \(3(400 - 100) = 3 \times 300 = 900.\) Correct.

Answers: (a) \(x = 44^\circ\); (b) Tank A \(= 1000\) litres, Tank B \(= 400\) litres.