Let me take a stab at (Q1). I find it better working for $\log\Gamma(z)$ rather than $\Gamma(z)$, since $\Gamma(z) = \exp( \log\Gamma(z))$.
Stirling formula reads as follows:
$$
\log\Gamma(z) \sim (z-\frac{1}{2}) \log(z) - z + \frac{1}{2} \log(2 \pi) + o(1)
$$
for $\vert z \vert \to \infty$ and $ \vert \arg(z) \vert < \pi - \epsilon$.
Notice that shifted formula is a simple consequence of the above:
$$
\begin{eqnarray}
\log\Gamma(z+a) &\sim& ( z+a -\frac{1}{2}) \log(z+a) - z - a + \frac{1}{2} \log(2 \pi) + o(1) \\
&\sim& ( z+a -\frac{1}{2}) \log(z) + ( z+a -\frac{1}{2}) \log(1+\frac{a}{z}) - z - a + \frac{1}{2} \log(2 \pi) + o(1) \\
&\sim& ( z+a -\frac{1}{2}) \log(z) + z \log(1+\frac{a}{z}) - z - a + \frac{1}{2} \log(2 \pi) + o(1) \\
&\sim& ( z+a -\frac{1}{2}) \log(z) - z + \frac{1}{2} \log(2 \pi) + o(1)
\end{eqnarray}
$$
Now
$$
\begin{eqnarray}
\log\Gamma(a z + b) + \log\Gamma(c z + d) & \sim &
( a z +b -\frac{1}{2}) \log(a z) - a z + \frac{1}{2} \log(2 \pi) + \\
& & ( c z +d -\frac{1}{2}) \log(c z) - c z + \frac{1}{2} \log(2 \pi) + o(1)
\end{eqnarray}
$$
Then it is a matter of recombining terms as $ ( \mathcal{A} + \mathcal{B} z) + (\kappa z + \rho -\frac{1}{2}) \log (\kappa z) - \kappa z + \frac{1}{2} \log(2 \pi) + o(1) $.
See the book of Paris and Kaminski to fill in the details.
Updated
Here we will study $\Gamma$ directly since its behavior is much more regular than $\zeta$ in the critical strip.
Since $\;\overline{\Gamma (z)}=\Gamma (\overline{z})\;$ a first idea is to use any formula for $\;\Gamma(z)\,$ and transform it with :
$$\Re\;\Gamma(z)=\frac{\Gamma(z)+\Gamma(\overline{z})}2,\quad\Im\;\Gamma(z)=\frac{\Gamma(z)-\Gamma(\overline{z})}{2\,i}$$
Let's apply this to some integral formulae like the definition $\;\displaystyle\Gamma(z):=\int_0^\infty e^{-t}\,t^{z-1}\,dt$ :
\begin{align}
\Re\;\Gamma(x+iy)&=\int_0^\infty e^{-t}\,\frac{t^{x+iy-1}+t^{x-iy-1}}2\,dt\\
&=\int_0^\infty e^{-t}\,t^{x-1}\,\frac{e^{iy\ln(t)}+e^{-iy\ln(t)}}2\,dt\\
&\tag{1}=\int_0^\infty e^{-t}\,t^{x-1}\,\cos(y\;\ln\,t)\;dt\\
\\\\
\Im\;\Gamma(x+iy)&=\int_0^\infty e^{-t}\,\frac{t^{x+iy-1}-t^{x-iy-1}}{2\,i}\,dt\\
\tag{2}&=\int_0^\infty e^{-t}\,t^{x-1}\,\sin(y\;\ln\,t)\;dt\\
\end{align}
The variant $\;\displaystyle\Gamma(z)=2\int_0^\infty e^{-t^2}\,t^{2z-1}\,dt\;$ returned :
\begin{align}
\tag{3}\Re\;\Gamma(x+iy)&=2\int_0^\infty e^{-t^2}\,t^{2x-1}\,\cos(2\,y\;\ln\,t)\;dt\\
\tag{4}\Im\;\Gamma(x+iy)&=2\int_0^\infty e^{-t^2}\,t^{2x-1}\,\sin(2\,y\;\ln\,t)\;dt\\
\end{align}
with faster convergence and the power of $t$ disappearing for $x=\frac 12$.
While the finite integral $\Gamma(z)=\int_0^1 (-\ln\,t)^{z-1}\,dt\;$ gives :
\begin{align}
\tag{5}\Re\;\Gamma(x+iy)&=\int_0^1 (-\ln\,t)^{x-1}\,\cos(y\;\ln(-\ln(t)))\;dt\\
\tag{6}\Im\;\Gamma(x+iy)&=\int_0^1 (-\ln\,t)^{x-1}\,\sin(y\;\ln(-\ln(t)))\;dt\\
\end{align}
We could try the same idea with infinite series and products from Abramowitz and Stegun but let's experiment another idea.
We may use $\;\displaystyle\Gamma(z)=|\Gamma(z)|\;e^{i\arg \Gamma(z)}\;$ to get :
\begin{align}
\Re\;\Gamma(z)&=\left|\Gamma(z)\right|\;\cos(\arg\;\Gamma(z))\\
\Im\;\Gamma(z)&=\left|\Gamma(z)\right|\;\sin(\arg\;\Gamma(z))\\
\end{align}
Let's obtain some moduli (absolute values) first :
As proposed by Kunnysan we may consider $\;\displaystyle\Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}\,$ applied to $z=\frac 12+iy\;$ and conclude using $\;\Gamma(\overline{z})=\overline{\Gamma (z)}\;$ that :
$$\tag{7}\left|\Gamma\left(\frac 12+iy\right)\right|=\sqrt{\frac{\pi}{\cosh(\pi\,y)}}$$
A more general formula from Mathias Lerch (but without explicit $\lambda(y)$) is for $0\le x\le 1$ :
$$\left|\Gamma\left(x+iy\right)\right|=\lambda(y)\frac{\Gamma(1+x)}{\sqrt{x^2+y^2}}\sqrt{\frac{\pi\;y}{\sin(\pi\,y)}}\quad\text{with}\ \ 1\le\lambda(y)\le\sqrt{1+y^2}$$
From A&S $6.1.25$ we have too :
$$\tag{8}\left|\Gamma\left(x+iy\right)\right|=\frac {\left|\Gamma(x)\right|}{\sqrt{\prod\limits_{n=0}^\infty 1+\frac {y^2}{(x+n)^2}}}$$
For numerical evaluation Stirling approximation may be useful :
$$\tag{9}\left|\Gamma\left(x+iy\right)\right|=\sqrt{2\pi}\,e^{-\pi|y|/2}\,|y|^{x-1/2}[1+r(x,y)]$$
with $|r(x,y)|\to 0$ uniformly for $x<K$ as $|y|\to\infty$.
(the error is around $1$% for $y=1$ and quickly decreasing with $y$ ; we will derive a similar but more precise formula in a moment)
While the argument of $\;\Gamma(x+iy)$ is given by A&S $6.1.27$ (if $\;x+iy\neq 0,-1,-2,\cdots$) :
$$\tag{10}\arg\;\Gamma(x+iy)=y\;\psi(x)+\sum_{n=0}^\infty\left[\frac y{x+n}-\arctan\frac y{x+n}\right]$$
with $\psi(x)=\dfrac{\Gamma'(x)}{\Gamma(x)}$ the psi or digamma function.
Observing that the argument is simply the imaginary part of $\,\ln\;\Gamma(z)\,$ we may use the asymptotic expansion $\ 6,1,40\,$ as $z\to\infty\,$ valid for $\,|\arg\;z|<\pi\,$ and with $B_n$ the Bernoulli numbers :
\begin{align}
\arg\,\Gamma(z)&\sim \Im\left[\left(z-\frac 12\right)\ln(z)-z+\frac 12\ln(2\pi)+\sum\limits_{m=1}^\infty \frac{B_{2m}}{2m\,(2m-1)}z^{1-2m}\right]\quad(11)\\
&\sim \Im\left(z-\frac 12\right)\Re\;\ln(z)+\Re\left(z-\frac 12\right)\Im\;\ln(z)-\Im\;z+\sum\limits_{m=1}^\infty \frac{B_{2m}}{2m\,(2m-1)}\Im\;{z^{1-2m}}\\
&\sim \frac{y\;\ln\bigl(x^2+y^2\bigr)}2+\left(x-\frac 12\right)\arg(x+iy)-y-\frac 1{12}\frac y{x^2+y^2}+\frac 1{360}\frac{3x^2y-y^3}{(x^2+y^2)^3}+\cdots\\
\end{align}
(for $\,\Re(x)>0$ you may replace $\,\arg(x+iy)\,$ by $\,\arctan\dfrac yx\,$ ; see arg for other quadrants)
The corresponding modulus is given by :
\begin{align}
|\Gamma(z)|&\sim \exp\;\Re\left[\left(z-\frac 12\right)\ln(z)-z+\frac 12\ln(2\pi)+\sum\limits_{m=1}^\infty \frac{B_{2m}\;z^{1-2m}}{2m\,(2m-1)}\right]\\
&\sim \sqrt{2\pi}\,\exp\left[\Re\left(z-\frac 12\right)\Re\;\ln(z)-\Im\left(z-\frac 12\right)\Im\;\ln(z)-\Re\;z+\sum\limits_{m=1}^\infty \frac{B_{2m}\;\Re\;{z^{1-2m}}}{2m\,(2m-1)}\right]\\
&\sim \sqrt{2\pi}\,\exp\left[\frac{\left(x-\frac 12\right)\;\ln\bigl(x^2+y^2\bigr)}2-y\;\arg(x+iy)-x+\sum\limits_{m=1}^\infty \frac{B_{2m}\;\Re\;{z^{1-2m}}}{2m\,(2m-1)}\right]\\
&\sim \sqrt{2\pi}\,\bigl(x^2+y^2\bigr)^{\frac x2-\frac 14}\exp\left[-x-y\,\arg(x+iy)+\frac 1{12}\frac x{x^2+y^2}-\frac 1{360}\frac{x^3-3xy^2}{(x^2+y^2)^3}+\cdots\right]\\
\end{align}
Combining these two formulas should provide accurate real and imaginary parts except near the negative axis.
Let's add that approximations for the $\Gamma$ function is a rather well studied subject and that the Lanczos approximation for example was often used in the critical strip.
Some illustrations to conclude :
The real part of $\Gamma$
The imaginary part of $\Gamma$
The modulus of $\Gamma$ (note the regularity of the surface at the right of the pole at $0$)
Best Answer
$1.$ Note that $\operatorname{Im}(\log \Gamma(z)) =\arg \Gamma(z)$.
$2.$ The general asymptotic expansion of the logarithm of the gamma function is $$\tag{1} \log \Gamma (z+h) \sim \left( {z+h - \frac{1}{2}} \right)\log z - z + \frac{1}{2}\log (2\pi ) + \sum\limits_{k = 2}^\infty {\frac{{B_{k}(h) }}{{k(k - 1)z^{k - 1} }}} , $$ as $z\to \infty$ in $|\arg z|\leq \pi -\delta$ ($<\pi$) with any fixed complex $h$. Here $B_k(h)$ denotes the Bernoulli polynomials. Taking $h=x$ and $z=\text{i}y$ we find $$\tag{2} \arg \Gamma (x+\text{i}y) \sim y\log y-y +\left(x-\frac{1}{2}\right)\!\frac{\pi }{2} + \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k B_{2k}(x) }}{{2k(2k - 1)y^{2k - 1} }}} , $$ as $y\to +\infty$ with fixed real $x$. Note that by the Schwarz reflection principle $\arg \Gamma (x-\text{i}y)=-\arg \Gamma (x+\text{i}y)$.
I shall add that there is a Stokes phenomenon happening on the imaginary axis for the logarithm of the Gamma function. This brings in exponentially small corrections to $(1)$, namely \begin{align*} \log \Gamma (z+h) \sim \left( {z+h - \frac{1}{2}} \right)\log z - z + \frac{1}{2}\log (2\pi ) & - \log (1 - {\rm e}^{\pm 2\pi {\rm i}(z + h)} )\\ & + \sum\limits_{k = 2}^\infty {\frac{{B_{k}(h) }}{{k(k - 1)z^{k - 1} }}} , \end{align*} as $z\to \infty$ in $\frac{\pi}{2}<\pm \arg z \leq \pi -\delta$ ($<\pi$) with any fixed complex $h$. Thus, $(2)$ is most useful when $x\geq 0$, whereas if $x<0$, one may use $$ \arg \Gamma (x+\text{i}y) \sim y\log y-y +\left(x-\frac{1}{2}\right)\!\frac{\pi }{2} +\! \sum\limits_{n = 1}^\infty {\frac{{\mathrm{e}^{ - 2\pi ny} }}{n}\sin (2\pi nx)} +\! \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k B_{2k}(x) }}{{2k(2k - 1)y^{2k - 1} }}} , $$ as $y\to +\infty$.