The answer is yes.

A first remark: since $A$ is positive-definite, $\lambda_i(A+B) \geq \lambda_i(B)$ for $i=1,2$. 
(to check this, use the formulas $\lambda_1(X) = \max_{\xi} \langle X\xi,\xi\rangle$ and $\lambda_2(X) = \min_{\xi} \langle X\xi,\xi\rangle$ where the min and max run over all unit vectors $\xi$. This formulas hold whenever $X$ is a symmetric $2 \times 2$ matrix.)

Your question is therefore whether $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ for any symmetric positive definite matrices $A$ and $B$, or equivalently $Tr( \sqrt{X X^{T} + Y Y^{T} } ) \leq Tr(\sqrt{X X^{T} })+Tr(\sqrt{Y Y^{T} })$ for any matrices $X$ and $Y$, where $X^{T}$ denotes the transpose of $X$, or hermitian tranpose if you work with complex matrices. This inequality is true in any dimension (not just 2), and it is just the triangle inequality for the [Schatten 1-norm][1] given by $\|X\|_1 = Tr(\sqrt{X X^{T} })$. The expression $Tr(\sqrt{X X^{T} + Y Y^{T} })$ is indeed the 1-norm of the matrix $\begin{pmatrix}X&Y \\\\ 0&0\end{pmatrix}$.

  [1]: http://en.wikipedia.org/wiki/Schatten_norm


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EDIT: I do not have the time to check the details, but the following should lead to an answer to your second question. It involves some computation. 

Denote by $\alpha_1\geq \alpha_2$ the eigenvalues of $A$, and the same with $\beta$ for $B$, and $\gamma$ for $A+B$. 

Given $\alpha$ and $\beta$, the possible values for $\gamma$ are described by Horn's inequalities. For $2 \times 2$ matrices, they are described by
$$\gamma_1 = Tr A + Tr B - \gamma_2,$$
$$\alpha_2+\beta_2 \leq\gamma_2 \leq \min(\alpha_1+\beta_2,\alpha_2+\beta_1).$$
But $\lambda_2(A+B)$ is $\gamma_2$ or $\gamma_1$, depending on the sign of $Tr(A+B)$.The possible values for $\lambda_2(A+B)$
thus form an interval $I$, and the endpoints of this interval correspond to the case when $A$ and $B$ commute.

Now, you can study how the LHS of your inequality varies as $\lambda_2(A+B)$ varies. Since $|\lambda_2(A+B)| \leq |\lambda_1(A+B)|$, you get that the derivative of the LHS has the same sign as the derivative of $|\sqrt{|\lambda_2(A+B)|} - \sqrt{|\lambda_2(B)|}|$. Its only local maximum on the interior of $I$ therefore corresponds to $\lambda_2(A+B)=0$ if $0\in I$. The global maximum of the LHS on $I$ is therefore reached at the endpoints of $I$, or at $\lambda_2(A+B)=0$ if $0\in I$.

But you know that on the endpoints, the inequality is verified since the matrices then commute. You are thus left to see whether
$\left|\sqrt{|Tr(A+B)|} - \sqrt{|\lambda_1(B)|}\right|+\sqrt{|\lambda_2(B)|} \leq \sqrt{|\alpha_1|}+\sqrt{|\alpha_2|}$ provided that $0$ belongs to $I$. 
Calculus should then allow you to find the maximum of the LHS on the domain defined by the inequalities $0 \in I$, $A$ being fixed. Then I expect that checking whether this maximum is less then $\sqrt{|\alpha_1|}+\sqrt{|\alpha_2|}$ should be again elementary calculus.