Although the answer of @SRKX is right on spot, I was already writing a solution along the lines of how you had specifically approached the problem. I think it might still be useful to you, so here it goes

The price of the chooser option, as seen of today $t=0$ is by definition \begin{align} V*0 &= \underbrace{e^{-r T*2}}*{\text{Payoff dicount factor}} \underbrace{\mathbb{E}\left[\ \ \underbrace{\max\left( \mathbb{E}[(S*{T*2}-K)^+ \vert \mathcal{F}*{T*1}], \mathbb{E}[(K-S*{T*2})^+ \vert \mathcal{F}*{T*1} ] \right)}*{\text{Expected payoff at $T*2$ as seen of $T*1$}} \ \ \vert \mathcal{F}*0 \ \ \right]}*{\text{Expected payoff at $T*2$ as seen of $t=0$}} \\ &= e^{-r T*2} \mathbb{E}*0\left[ \max\left( \mathbb{E}*{T*1}[(S*{T*2}-K)^+], \mathbb{E}*{T*1}[(K-S*{T*2})^+] \right) \right] \end{align} If you're not familiar with the notation $\mathcal{F}*t$ used for filtrations, you can interpret it as "all the information we know at time $t$". The notation $\mathbb{E}*t[.]$ simply figures that the expectation is taken conditionally on the knowledge of $\mathcal{F}*t$. Naturally all of these expectations are taken under the risk-neutral measure $\mathbb{Q}$.

By definition, we also have that the price of European call/put options is given by $$ C(T*1,S*{T*1};K,(T*2-T*1)) = e^{-r(T*2-T*1)} \mathbb{E}*{T*1}[(S*{T*2}-K)^+] \tag{def 1} := C*{12} $$ $$ P(T*1,S*{T*1};K,(T*2-T*1)) = e^{-r(T*2-T*1)}\mathbb{E}*{T*1}[(K-S*{T*2})^+] \tag{def 2} := P*{12} $$ where $C(t,S*t;K,\tau)$ (resp. $P(t,S*t;K,\tau)$) denotes the price of a European call (resp. put) option as seen of time $t$, given the spot value $S_t$, the strike price $K$ and the time to expiry $\tau$.

Therefore, $$ V*0 = e^{-r T*2} \mathbb{E}*0\left[ \max\left( \frac{C*{12}}{e^{-r(T*2-T*1)}}, \frac{P*{12}}{e^{-r(T*2-T*1)}} \right) \right]$$ Yet by call-put parity: $$ C*{12} - P*{12} = e^{-r(T*2-T*1)}( S*1 e^{(r-q)(T*2-T*1)} - K ) $$ so that we can further write (similarly to what you did) \begin{align} V*0 &= e^{-r T*2} \mathbb{E}*0\left[ \max\left( \frac{C*{12}}{e^{-r(T*2-T*1)}}, \frac{C*{12}}{e^{-r(T*2-T*1)}} - (S*1e^{(r-q)(T*2-T*1)} - K) \right) \right] \\ &= e^{-r T*2} \mathbb{E}*0\left[ \left( \frac{C*{12}}{e^{-r(T*2-T*1)}} + \max\left( 0, K - S*1e^{(r-q)(T*2-T*1)} \right) \right) \right] \\ &= \mathbb{E}*0\left[ e^{-rT*1} C*{12} \right] + \mathbb{E}*0\left[ e^{-rT*2} \max\left( 0, K - S*1e^{(r-q)(T*2-T*1)} \right) \right] \tag{1} \end{align}

Now using $(\text{def } 1)$ the first term of $(1)$ becomes: $$ \mathbb{E}*0 \left[ e^{-rT*1} C*{12} \right] = \mathbb{E}*0 \left[ e^{-rT*2} \mathbb{E}*{T*1}[(S*{T*2}-K)^+] \right] = C(0,S*0;K,T_2)$$ by the tower property of conditional expectations.

Similarly, the second term of $(1)$ can on the other hand be expressed as: \begin{align} \mathbb{E}*0\left[ e^{-rT*2} \max\left( 0, K - S*1e^{(r-q)(T*2-T*1)} \right) \right] &= \mathbb{E}*0\left[ \max\left( 0, Ke^{-rT*2} - S*1e^{-rT*1-q(T*2-T*1)} \right) \right] \\ &= e^{-q(T*2-T*1)} \mathbb{E}*0\left[ e^{-r{T*1}} \max\left( 0, Ke^{-(r-q)(T*2-T*1)} - S*1 \right) \right] \\ &= e^{-q(T*2-T*1)} P(0,S*0; Ke^{-(r-q)(T*2-T*1)}, T*1) \end{align} So that $(1)$ becomes $$ V*0 = C(0,S*0;K,T*2) + \underbrace{e^{-q(T*2-T*1)}}*{= 0.9778} P(0,S*0; \underbrace{Ke^{-(r-q)(T*2-T*1)}}*{= 62.1085}, T*1) $$ hence a $T*2$ call struck at $K$ + 0.9778 units of a $T_1$ put with adjusted strike 62.1085.