Answer
a) $1.32$
b) $1.25$
Work Step by Step
\nea) We are told that the gas undergoes a compression-adiabatic process and then $V_f=\frac{1}{2}V_i$, and $P_f=2.5P_i$.
For an adiabatic process,
$$P_iV_i^\gamma=P_fV_f^\gamma$$
Plugging the known;
$$ \color{red}{\bf\not} P_iV_i^\gamma=(2.5 \color{red}{\bf\not} P_i)\left(\frac{1}{2}V_i\right)^\gamma$$
Thus,
$$\left(\dfrac{ \color{red}{\bf\not} V_i}{\frac{1}{2} \color{red}{\bf\not} V_i}\right)^\gamma=2.5$$
$$(2)^\gamma=2.5$$
Taking the logarithm base-2 for both sides;
$$\log_2(2)^\gamma=\log_2(2.5)$$
$$\gamma\;\;\overbrace{\log_2(2)}^{=1}=\log_2(2.5)$$
$$\gamma =\log_2(2.5)=\color{red}{\bf 1.32}$$
Noting that you can take the natural logarithm as well and get the same result.
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b) We know, for an adiabatic process, that
$$T_iV_i^{\gamma-1}=T_fV_f^{\gamma-1}$$
Hence,
$$\dfrac{T_f}{T_i}=\dfrac{V_i^{\gamma-1}}{V_f^{\gamma-1}}=\left(\dfrac{ \color{red}{\bf\not} V_i}{\frac{1}{2} \color{red}{\bf\not} V_i}\right)^{\gamma-1}$$
$$\dfrac{T_f}{T_i}=2^{\gamma-1}$$
Plugging from above;
$$\dfrac{T_f}{T_i}=2^{(1.32-1)}=\color{red}{\bf 1.25}$$
