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$\newcommand\PP{\mathcal P}\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ga{\gamma}$1. The answer to Question 1 is yes. Indeed, suppose that a sequence of probability measures $\mu_n$ with densities $\rho_n$ converges in $\tau$ to a probability measure $\mu$ with density $\rho$. For real $\si>0$, let $\ga_\si$ be the Gaussian measure over $\R^d$ with mean $0$ and covariance matrix $\si I_d$, where $I_d$ is the identity matrix. Let $f_\si$ be the density of $\ga_\si$.

Then for each $\si>0$ the density $f_\si *\rho_n$ of the probability measure $\ga_\si *\mu_n$ converges pointwise to the density $f_\si *\rho$ of the probability measure $\ga_\si *\mu$ (this follows because $f_\si$ is a bounded continuous function). So, by the Fatou lemma, $$\liminf_n\H(f_\si *\rho_n)\ge \H(f_\si *\rho). \tag{1}\label{1}$$

The function $h\colon[0,\infty)\to\R$ given by $h(t):=t\log t$ for $t>0$ with $h(0):=0$ is convex. So, by Jensen's inequality, $$\H(\rho_n)\ge\H(f_\si *\rho_n). \tag{2}\label{2}$$

Also, $f_\si *\rho\to\rho$ almost everywhere as $\si\downarrow0$, and the function $h$ is bounded from below. So, again by the Fatou lemma, $$\liminf_{\si\downarrow0}\H(f_\si *\rho)\ge \H(\rho). \tag{3}\label{3}$$

It follows by \eqref{2},\eqref{1}, and \eqref{3} that $\liminf_n\H(\rho_n)\ge \H(\rho)$. So, $\H$ is l.s.c. in $\tau$. $\quad\Box$

2. The answer to Question 2 is no: As in the comment by Kostya_I, let $\rho_n(x):=\rho(x-n)$.

$\newcommand\PP{\mathcal P}\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ga{\gamma}$1. The answer to Question 1 is no. Indeed, let $d=1$ and $$\rho_n:=c_n p_n,$$ where $$p_n(x):=f(x)+\frac{1(e^n<x<e^{e^n})}{x\ln^2 x}$$ for natural $n$ and real $x$, $f$ is the standard normal pdf, and $c_n:=1/\int_\R p_n\to1$.

Then $\rho_n\to\rho:=f$ pointwise (as $n\to\infty$), so that the probability measures with densities $\rho_n$ converge in total variation and hence in $\tau$ to the probability measure with density $\rho$.

On the other hand (assuming that your $\log$ is $\ln$), $$\H(\rho_n)\sim-\int_{e^n}^{e^{e^n}}\frac{dx}{x\ln x}\sim -n\to-\infty,$$ whereas $\H(\rho)\in\R$. $\quad\Box$

2. The answer to Question 2 is no: As in the comment by Kostya_I, let $\rho_n(x):=\rho(x-n)$.

$\newcommand\PP{\mathcal P}\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ga{\gamma}$1. The answer to Question 1 is no. Indeed, let $d=1$ and $$\rho_n:=c_n p_n,$$ where $$p_n(x):=f(x)+\frac{1(e^n<x<e^{e^n})}{x\ln^2 x}$$ for natural $n$ and real $x$, $f$ is the standard normal pdf, and $c_n:=1/\int_\R p_n\to1$.

Then $\rho_n\to\rho:=f$ pointwise (as $n\to\infty$), so that the probability measures with densities $\rho_n$ converge in total variation and hence in $\tau$ to the probability measure with density $\rho$.

On the other hand (assuming that your $\log$ is $\ln$), $$\H(\rho_n)\sim-\int_{e^n}^{e^{e^n}}\frac{dx}{x\ln x}\sim -n\to-\infty,$$ whereas $\H(\rho)\in\R$. $\quad\Box$

2. The answer to Question 2 is no: As in the comment by Kostya_I, let $\rho_n(x):=\rho(x-n)$.

$\newcommand\PP{\mathcal P}\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ga{\gamma}$1. The answer to Question 1 is yes. Indeed, suppose that a sequence of probability measures $\mu_n$ with densities $\rho_n$ converges in $\tau$ to a probability measure $\mu$ with density $\rho$. For real $\si>0$, let $\ga_\si$ be the Gaussian measure over $\R^d$ with mean $0$ and covariance matrix $\si I_d$, where $I_d$ is the identity matrix. Let $f_\si$ be the density of $\ga_\si$.

Then for each $\si>0$ the density $f_\si *\rho_n$ of the probability measure $\ga_\si *\mu_n$ converges pointwise to the density $f_\si *\rho$ of the probability measure $\ga_\si *\mu$ (this follows because $f_\si$ is a bounded continuous function). So, by the Fatou lemma, $$\liminf_n\H(f_\si *\rho_n)\ge \H(f_\si *\rho). \tag{1}\label{1}$$

The function $h\colon[0,\infty)\to\R$ given by $h(t):=t\log t$ for $t>0$ with $h(0):=0$ is convex. So, by Jensen's inequality, $$\H(\rho_n)\ge\H(f_\si *\rho_n). \tag{2}\label{2}$$

Also, $f_\si *\rho\to\rho$ almost everywhere as $\si\downarrow0$, and the function $h$ is bounded from below. So, again by the Fatou lemma, $$\liminf_{\si\downarrow0}\H(f_\si *\rho)\ge \H(\rho). \tag{3}\label{3}$$

It follows by \eqref{2},\eqref{1}, and \eqref{3} that $\liminf_n\H(\rho_n)\ge \H(\rho)$. So, $\H$ is l.s.c. in $\tau$. $\quad\Box$

2. The answer to Question 2 is no: As in the comment by Kostya_I, let $\rho_n(x):=\rho(x-n)$.

$\newcommand\PP{\mathcal P}\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ga{\gamma}$1. The answer to Question 1 is no. Indeed, let $d=1$ and $$\rho_n:=c_n p_n,$$ where $$p_n(x):=f(x)+\frac{1(e^n<x<e^{e^n})}{x\ln^2 x}$$ for natural $n$ and real $x$, $f$ is the standard normal pdf, and $c_n:=1/\int_\R p_n\to1$.

Then $\rho_n\to\rho:=f$ pointwise (as $n\to\infty$), so that the corresponding probability measures converge in total variation and hence in $\tau$.

On the other hand, $$\H(\rho_n)\sim-\int_{e^n}^{e^{e^n}}\frac{dx}{x\ln x}\sim n\to-\infty,$$ whereas $\H(\rho)\in\R$. $\quad\Box$

2. The answer to Question 2 is no: As in the comment by Kostya_I, let $\rho_n(x):=\rho(x-n)$.

$\newcommand\PP{\mathcal P}\newcommand\H{\mathcal H}\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ga{\gamma}$1. The answer to Question 1 is no. Indeed, let $d=1$ and $$\rho_n:=c_n p_n,$$ where $$p_n(x):=f(x)+\frac{1(e^n<x<e^{e^n})}{x\ln^2 x}$$ for natural $n$ and real $x$, $f$ is the standard normal pdf, and $c_n:=1/\int_\R p_n\to1$.

Then $\rho_n\to\rho:=f$ pointwise (as $n\to\infty$), so that the probability measures with densities $\rho_n$ converge in total variation and hence in $\tau$ to the probability measure with density $\rho$.

On the other hand (assuming that your $\log$ is $\ln$), $$\H(\rho_n)\sim-\int_{e^n}^{e^{e^n}}\frac{dx}{x\ln x}\sim -n\to-\infty,$$ whereas $\H(\rho)\in\R$. $\quad\Box$

2. The answer to Question 2 is no: As in the comment by Kostya_I, let $\rho_n(x):=\rho(x-n)$.