Sampling out the scenario we get
${HH,HT,TH,TT}$ each with probability $\frac{1}{4}$
$P(X=no. of heads)$
$E[X]\Rightarrow 2*\frac{1}{4} + 1*\frac{1}{4}+1*\frac{1}{4}+0*\frac{1}{4}$
$\Rightarrow \frac{2}{4}+\frac{2}{4}\Rightarrow 1$
$E[X]=1$; $E^{2}[X]\Rightarrow 1$
Now calculating for $E[X^{2}]$
$E[X^{2}]\Rightarrow 2^{2}*\frac{1}{4}+1^{2}*\frac{1}{4}+1^{2}*\frac{1}{4}+0^{2}*\frac{1}{4}$
$\Rightarrow 1+\frac{1}{2}\Rightarrow 1.5$
so $E[X^{2}]$$=1.5$