Answer
a) $h(1)=b+c+1$
b) $h(b+1)=2 b^2+3 b+c+1$
Work Step by Step
(a) $h(1)=(1)^2+b(1)^2+c=b+c+1$
(b) Substituting $b+1$ for $x$ in the formula for $h(x)$ :
$$
\begin{aligned}
h(b+1) & =(b+1)^2+b(b+1)+c \\
& =\left(b^2+2 b+1\right)+b^2+b+c \\
& =2 b^2+3 b+c+1
\end{aligned}
$$
