Let the initial state be the moment the block first contacts the spring (with speed $v$), and the final state be when the block momentarily stops (speed is zero, compression is $x$). The system includes the block, spring, and floor.

[A1] The work done by the spring force, $W_s$, as it is compressed by a distance $x$ is the negative of the change in its potential energy.

$$W_s = -\Delta U_s = -( \frac{1}{2}kx^2 - 0 ) = -\frac{1}{2}kx^2$$

[A2] The increase in thermal energy, $\Delta E_{th}$, is equal to the energy dissipated by friction. This is the magnitude of the work done by the friction force, $W_f$. The friction force is $f_k = \mu_k N = \mu_k mg$ and it acts over a distance $x$.

$$W_f = -f_k x = -\mu_k mgx$$ $$\Delta E_{th} = -W_f = \mu_k mgx$$

[A3] We apply the work-energy theorem to the block. The net work done on the block equals its change in kinetic energy, $\Delta K = W_{net}$. The initial kinetic energy is $K_i = \frac{1}{2}mv^2$ and the final kinetic energy is $K_f=0$. The net work is the sum of the work done by the spring and friction.

$$\Delta K = W_s + W_f$$ $$0 - \frac{1}{2}mv^2 = -\frac{1}{2}kx^2 - \mu_k mgx$$

Multiplying by -2 and solving for $v^2$:

$$mv^2 = kx^2 + 2\mu_k mgx$$ $$v = \sqrt{\frac{kx^2 + 2\mu_k mgx}{m}} = \sqrt{\frac{k}{m}x^2 + 2\mu_k gx}$$