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} V0 &= \underbrace{e^{-r T2}}{\text{Payoff dicount factor}} \underbrace{\mathbb{E}\left[\ \ \underbrace{\max\left( \mathbb{E}[(S{T2}-K)^+ \vert \mathcal{F}{T1}], \mathbb{E}[(K-S{T2})^+ \vert \mathcal{F}{T1} ] \right)}{\text{Expected payoff at $T2$ as seen of $T1$}} \ \ \vert \mathcal{F}0 \ \ \right]}{\text{Expected payoff at $T2$ as seen of $t=0$}} \\ &= e^{-r T2} \mathbb{E}0\left[ \max\left( \mathbb{E}{T1}[(S{T2}-K)^+], \mathbb{E}{T1}[(K-S{T2})^+] \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 $$C(T1,S{T1};K,(T2-T1)) = e^{-r(T2-T1)} \mathbb{E}{T1}[(S{T2}-K)^+] \tag{def 1} := C{12}$$ $$P(T1,S{T1};K,(T2-T1)) = e^{-r(T2-T1)}\mathbb{E}{T1}[(K-S{T2})^+] \tag{def 2} := P{12}$$ where $C(t,St;K,\tau)$ (resp. $P(t,St;K,\tau)$) 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, $$V0 = e^{-r T2} \mathbb{E}0\left[ \max\left( \frac{C{12}}{e^{-r(T2-T1)}}, \frac{P{12}}{e^{-r(T2-T1)}} \right) \right]$$ Yet by call-put parity: $$C{12} - P{12} = e^{-r(T2-T1)}( S1 e^{(r-q)(T2-T1)} - K )$$ so that we can further write (similarly to what you did) \begin{align} V0 &= e^{-r T2} \mathbb{E}0\left[ \max\left( \frac{C{12}}{e^{-r(T2-T1)}}, \frac{C{12}}{e^{-r(T2-T1)}} - (S1e^{(r-q)(T2-T1)} - K) \right) \right] \\ &= e^{-r T2} \mathbb{E}0\left[ \left( \frac{C{12}}{e^{-r(T2-T1)}} + \max\left( 0, K - S1e^{(r-q)(T2-T1)} \right) \right) \right] \\ &= \mathbb{E}0\left[ e^{-rT1} C{12} \right] + \mathbb{E}0\left[ e^{-rT2} \max\left( 0, K - S1e^{(r-q)(T2-T1)} \right) \right] \tag{1} \end{align}

Now using $(\text{def } 1)$ the first term of $(1)$ becomes: $$\mathbb{E}0 \left[ e^{-rT1} C{12} \right] = \mathbb{E}0 \left[ e^{-rT2} \mathbb{E}{T1}[(S{T2}-K)^+] \right] = C(0,S0;K,T_2)$$ 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^{-rT2} \max\left( 0, K - S1e^{(r-q)(T2-T1)} \right) \right] &= \mathbb{E}0\left[ \max\left( 0, Ke^{-rT2} - S1e^{-rT1-q(T2-T1)} \right) \right] \\ &= e^{-q(T2-T1)} \mathbb{E}0\left[ e^{-r{T1}} \max\left( 0, Ke^{-(r-q)(T2-T1)} - S1 \right) \right] \\ &= e^{-q(T2-T1)} P(0,S0; Ke^{-(r-q)(T2-T1)}, T1) \end{align} So that $(1)$ becomes $$V0 = C(0,S0;K,T2) + \underbrace{e^{-q(T2-T1)}}{= 0.9778} P(0,S0; \underbrace{Ke^{-(r-q)(T2-T1)}}{= 62.1085}, T1)$$ hence a $T2$ call struck at $K$ + 0.9778 units of a $T_1$ put with adjusted strike 62.1085.