Distinguish a transformed observation from two independent observations

$p+\frac12+\frac14+\frac18+\frac1{16}=1$, so $p=1/16$. Use parts(i)–(ii) as a self-read revision before focusing on the transformations and independent observations, following the Teacher guidance.

The completed probabilities at $0,1,2,4,8$ are $1/16,1/2,1/4,1/8,1/16$. Thus $E(X)=0(\frac1{16})+1(\frac12)+2(\frac14)+4(\frac18)+8(\frac1{16})=2$. Also $E(X^2)=0^2(\frac1{16})+1^2(\frac12)+2^2(\frac14)+4^2(\frac18)+8^2(\frac1{16})=15/2$. Therefore $\operatorname{Var}(X)=E(X^2)-[E(X)]^2=15/2-2^2=7/2$.

$E(2X-3)=2E(X)-E(3)=2(2)-3=1$. This is the Teacher completion.

$\operatorname{Var}(2X-3)=2^2\operatorname{Var}(X)+\operatorname{Var}(3)=4(7/2)+0=14$. The source explicitly includes the zero variance of the constant. This is a constant-shift rule; it does not justify adding variances for arbitrary dependent variables.

$E(X_1+X_2)=E(X_1)+E(X_2)=2+2=4$.

The disjoint ordered cases are $(0,2),(1,1),(2,0)$. Independence gives $P(X_1+X_2=2)=P(X_1=0)P(X_2=2)+P(X_1=1)P(X_2=1)+P(X_1=2)P(X_2=0)=(\frac1{16})(\frac14)+(\frac12)(\frac12)+(\frac14)(\frac1{16})=9/32$.