The specifications of transistor 2N525 are

$P_{D(max)}=225mW$

$h_{fe}=45(typ)$

$h_{c}=1.4 K\Omega \ \ \ \ h_{re}=3.2\times 10^{-4}, \ \ \ \ \ \ \ h_{oe}=25\mu mho$

i)Calculation of $R_{C}$

The voltage gain is given as

$A_{v}=\frac{h_{fe}R_{c}}{h_{fe}+\Delta hR_{c}}$

and $\Delta h=h_{ie}h_{oe}-h_{fe}h_{re}$

=$(1.4k\times 25 \mu=mho)-(45\times 3.2 \times 10^{-4})$

$\Delta h=0.0206$

40=$\frac{45\times R_{c}}{1.4k+0.0206\times R_{C}}$ …