$P(\text{win on a throw})=\frac{2}{6}=\frac{1}{3}$, $P(\text{lose})=\frac{2}{3}$.

The player stops as soon as he wins or after 3 throws.

• Win on throw 1: prob $=\frac{1}{3}$; gain $=+100$.
• Win on throw 2 (lost throw 1): prob $=\frac{2}{3}\cdot\frac{1}{3}=\frac{2}{9}$; net $=-50+100=+50$.
• Win on throw 3 (lost throws 1,2): prob $=\frac{2}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}=\frac{4}{27}$; net $=-100+100=0$.
• Lose all 3 throws: prob $=\left(\frac{2}{3}\right)^3=\frac{8}{27}$; net $=-150$.

$E = \frac{1}{3}(100)+\frac{2}{9}(50)+\frac{4}{27}(0)+\frac{8}{27}(-150)$

$=\frac{100}{3}+\frac{100}{9}+0-\frac{1200}{27}$

$=\frac{900}{27}+\frac{300}{27}-\frac{1200}{27}=\frac{0}{27}=\mathbf{0}$

Chloroxylenol contains a phenolic –OH group → Ferric chloride test (R). Norethindrone has a primary amine group → Carbylamine test (P). Sulphapyridine contains a C=C double bond (unsaturation) → Bayer's test (S). Penicillin is a carboxylic acid → Sodium hydrogencarbonate test (Q). So: $A \to R,\ B \to P,\ C \to S,\ D \to Q$.

Apply the addition formula: $\tan^{-1}(2x)+\tan^{-1}(3x)=\tan^{-1}\!\left(\dfrac{5x}{1-6x^2}\right)$ when $6x^2<1$.

Setting equal to $\pi/4$: $\dfrac{5x}{1-6x^2}=1 \Rightarrow 5x=1-6x^2 \Rightarrow 6x^2+5x-1=0$

$x=\dfrac{-5\pm\sqrt{25+24}}{12}=\dfrac{-5\pm7}{12}$

$x=\dfrac{2}{12}=\dfrac{1}{6}$ or $x=\dfrac{-12}{12}=-1$.

Since $x\geq0$, only $x=\dfrac{1}{6}$ is valid. Check: $6\cdot(1/6)^2=1/6<1$ ✓. So $A=\{1/6\}$ — a singleton.