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(1) (i) By APPlying KVL,\begin{array}{l}V_{C C}-I_{C} R_{C}=V_{\text {out }} \\V_{\text {in }}-I_{B} R_{B}-V_{B E}=0\end{array}As we know Voltage gain \Rightarrow A_{V}=\frac{V_{\text {out }}}{V_{\text {in }}}.\begin{array}{l}V_{\text {out }}=V_{C C}-I_{C} R_{C} \\V_{\text {in }}=V_{B E}+I_{B} R_{B} \\\therefore A_{V}=\left[\frac{V_{C C}-I_{C} R_{C}}{V_{B E}+I_{B} R_{B}}\right]\end{array}Now we have to find out expression for input and output resistance.\begin{array}{l}\text { Output Resistance }\left(R_{C}\right)=\frac{V_{C C}-V_{\text {out }}}{I_{C}} \\\text { Input Resistance }\left(R_{B}\right)=\frac{V_{\text {in }}-V_{B E}}{I_{B}}\end{array}(ii) Now for 2^{\text {nd }} part by applying K V L, writing KVL for input side,V_{\text {in }}-I_{B} R_{B}-V_{B E}=0Writing KVL for outputside,\begin{array}{l} V_{\text {out }}-I_{C} R_{C}-V_{c C}-V_{B}=0 \\\therefore V_{i n}=V_{B E}+I_{B} R_{B} \\V_{\text {out }}=V_{B}+V_{c c}+I_{C} R_{C}\end{array}So, A_{V}=\frac{V_{\text {out }}}{V_{\text {in }}}=\left[\frac{V_{B}+V_{C C}+I_{C} R_{C}}{V_{B E}+I_{B} R_{B}}\right]Now let us find R_{B}=\frac{V_{\text {in }}-V_{B E}}{I_{B}}\text { similarly } R_{C}=\frac{V_{\text {out }}-V_{B}-V_{C C}}{I_{C}} ...